RS Aggarwal Solutions for Class 11 Chapter 27 Limits

Question 1. Evaluate \( \lim_{x \to 2} (5 - x) \)
Answer: Using the formula \( \lim_{x \to a} f(x) = f(a) \), we substitute x = 2 into the expression (5 - x). This gives us 5 - 2 = 3. Therefore, \( \lim_{x \to 2} (5 - x) = 3 \).
In simple words: When x approaches 2, the value 5 minus x gets closer to 3.

Exam Tip: For continuous functions (like polynomials), direct substitution works perfectly - just plug in the value that x approaches.

Question 2. Evaluate \( \lim_{x \to 1} (6x^2 - 4x + 3) \)
Answer: Using the formula \( \lim_{x \to a} f(x) = f(a) \), we substitute x = 1 into the polynomial 6x² - 4x + 3. This gives us 6(1)² - 4(1) + 3 = 6 - 4 + 3 = 5. Therefore, \( \lim_{x \to 1} (6x^2 - 4x + 3) = 5 \).
In simple words: Replace x with 1 in the expression and calculate the result.

Exam Tip: Always verify your arithmetic when combining positive and negative terms - check that 6 - 4 + 3 really equals 5.

Question 3. Evaluate \( \lim_{x \to 1} \left( \frac{x^2 + 9}{x + 3} \right) \)
Answer: Using the formula \( \lim_{x \to a} f(x) = f(a) \), we substitute x = 3 into the fraction. The numerator becomes 3² + 9 = 9 + 9 = 18, and the denominator becomes 3 + 3 = 6. Therefore, \( \lim_{x \to 3} \frac{x^2 + 9}{x + 3} = \frac{18}{6} = 3 \).
In simple words: Plug x = 3 into both the top and bottom of the fraction, then simplify.

Exam Tip: Ensure the denominator is not zero before using direct substitution - if it is zero, you need a different method.

Question 4. Evaluate \( \lim_{x \to 3} \left( \frac{x^2 - 4x}{x - 2} \right) \)
Answer: Using the formula \( \lim_{x \to a} f(x) = f(a) \), we substitute x = 3 into the fraction. The numerator becomes 3² - 4(3) = 9 - 12 = -3, and the denominator becomes 3 - 2 = 1. Therefore, \( \lim_{x \to 3} \frac{x^2 - 4x}{x - 2} = \frac{-3}{1} = -3 \).
In simple words: Calculate the numerator and denominator separately by substituting x = 3, then divide the results.

Exam Tip: Take care with negative values - a negative numerator divided by a positive denominator gives a negative result.

Question 5. Evaluate \( \lim_{x \to 5} \left( \frac{x^2 - 25}{x - 5} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We factor the numerator as a difference of squares: \( x^2 - 25 = (x + 5)(x - 5) \). Substituting this factorization: \( \lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} \frac{(x + 5)(x - 5)}{x - 5} = \lim_{x \to 5} (x + 5) = 5 + 5 = 10 \).
In simple words: When you get 0/0, factor the numerator and cancel the common factor with the denominator before substituting.

Exam Tip: Recognize the difference of squares pattern a² - b² = (a + b)(a - b) to factor and cancel efficiently.

Question 6. Evaluate \( \lim_{x \to 1} \left( \frac{x^3 - 1}{x - 1} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We use the factorization \( x^3 - 1 = (x - 1)(x^2 + x + 1) \). Substituting: \( \lim_{x \to 1} \frac{x^3 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1} = \lim_{x \to 1} (x^2 + x + 1) = 1 + 1 + 1 = 3 \).
In simple words: Factor the cubic expression using the difference of cubes formula, cancel the (x - 1) terms, then substitute.

Exam Tip: Remember that x³ - y³ = (x - y)(x² + xy + y²) - this pattern appears frequently in limit problems.

Question 9. Evaluate \( \lim_{x \to 3} \left( \frac{x^2 - 4x + 3}{x^2 - 2x - 3} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We factor both numerator and denominator. The numerator x² - 4x + 3 factors as (x - 3)(x - 1). The denominator x² - 2x - 3 factors as (x - 3)(x + 2). Substituting: \( \lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3} = \lim_{x \to 3} \frac{(x - 3)(x - 1)}{(x - 3)(x + 2)} = \lim_{x \to 3} \frac{x - 1}{x + 2} = \frac{3 - 1}{3 + 2} = \frac{2}{5} \).
In simple words: Factor both the numerator and denominator, cancel the common (x - 3) factor, then evaluate by substitution.

Exam Tip: Always factor quadratic expressions by finding two numbers that multiply to give the constant term and add to give the middle coefficient.

Question 10. Evaluate \( \lim_{x \to \frac{1}{2}} \left( \frac{4x^2 - 1}{2x - 1} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We factor the numerator as a difference of squares: \( 4x^2 - 1 = (2x)^2 - 1^2 = (2x + 1)(2x - 1) \). Substituting: \( \lim_{x \to \frac{1}{2}} \frac{4x^2 - 1}{2x - 1} = \lim_{x \to \frac{1}{2}} \frac{(2x + 1)(2x - 1)}{2x - 1} = \lim_{x \to \frac{1}{2}} (2x + 1) = 2 \cdot \frac{1}{2} + 1 = 1 + 1 = 2 \).
In simple words: Recognize 4x² - 1 as a difference of squares, factor it, cancel the common term (2x - 1), then substitute.

Exam Tip: When the limit point is a fraction, be especially careful with arithmetic - compute 2(1/2) + 1 step by step to avoid errors.

Question 11. Evaluate \( \lim_{x \to 4} \left( \frac{x^3 - 64}{x^2 - 16} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We factor the numerator using the difference of cubes: \( x^3 - 64 = x^3 - 4^3 = (x - 4)(x^2 + 4x + 16) \). We factor the denominator as a difference of squares: \( x^2 - 16 = (x + 4)(x - 4) \). Substituting: \( \lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16} = \lim_{x \to 4} \frac{(x - 4)(x^2 + 4x + 16)}{(x + 4)(x - 4)} = \lim_{x \to 4} \frac{x^2 + 4x + 16}{x + 4} = \frac{16 + 16 + 16}{4 + 4} = \frac{48}{8} = 6 \).
In simple words: Apply the difference of cubes formula to the numerator and the difference of squares formula to the denominator, cancel (x - 4), then substitute x = 4.

Exam Tip: When both numerator and denominator factor, cancel all common terms carefully before substituting to avoid unnecessary calculation.

Question 12. Evaluate \( \lim_{x \to 2} \left( \frac{x^5 - 32}{x^3 - 8} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We recognize both terms using the formula \( \frac{x^m - y^m}{x - y} = my^{m-1} \) where applicable. For the numerator: \( x^5 - 32 = x^5 - 2^5 \). For the denominator: \( x^3 - 8 = x^3 - 2^3 \). Factoring: \( x^5 - 2^5 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16) \) and \( x^3 - 2^3 = (x - 2)(x^2 + 2x + 4) \). Substituting: \( \lim_{x \to 2} \frac{x^5 - 32}{x^3 - 8} = \lim_{x \to 2} \frac{(x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)}{(x - 2)(x^2 + 2x + 4)} = \lim_{x \to 2} \frac{x^4 + 2x^3 + 4x^2 + 8x + 16}{x^2 + 2x + 4} = \frac{16 + 16 + 16 + 16 + 16}{4 + 4 + 4} = \frac{80}{12} = \frac{20}{3} \).
In simple words: Factor both numerator and denominator by removing the common (x - 2) factor, simplify the resulting fraction, then substitute x = 2.

Exam Tip: For higher-degree expressions like x⁵ and x³, use the general difference of powers factorization systematically to avoid mistakes.

Question 13. Evaluate \( \lim_{x \to a} \left( \frac{x^{\frac{5}{2}} - a^{\frac{5}{2}}}{x - a} \right) \)
Answer: Using the general limit formula \( \lim_{x \to a} \frac{x^m - y^m}{x - y} = my^{m-1} \), we apply it with m = 5/2 and y = a. Therefore, \( \lim_{x \to a} \frac{x^{\frac{5}{2}} - a^{\frac{5}{2}}}{x - a} = \frac{5}{2} a^{\frac{5}{2} - 1} = \frac{5}{2} a^{\frac{3}{2}} \).
In simple words: Use the power limit formula: bring down the exponent and reduce it by 1, multiply by the constant raised to the new power.

Exam Tip: This formula works for any real exponent m, including fractions - memorizing this pattern saves time on many limit problems.

Question 14. Evaluate \( \lim_{x \to a} \left\{ \frac{(x + 2)^{\frac{5}{3}} - (a + 2)^{\frac{5}{3}}}{x - a} \right\} \)
Answer: Let u = x + 2 and b = a + 2. As x approaches a, u approaches b. The limit becomes \( \lim_{u \to b} \frac{u^{\frac{5}{3}} - b^{\frac{5}{3}}}{u - b} \). Using the limit formula \( \lim_{u \to b} \frac{u^m - b^m}{u - b} = mb^{m-1} \) with m = 5/3: \( \lim_{u \to b} \frac{u^{\frac{5}{3}} - b^{\frac{5}{3}}}{u - b} = \frac{5}{3} b^{\frac{5}{3} - 1} = \frac{5}{3} b^{\frac{2}{3}} = \frac{5}{3} (a + 2)^{\frac{2}{3}} \).
In simple words: Substitute u = x + 2 to simplify the expression, apply the power limit formula with the new variable, then convert back using (a + 2).

Exam Tip: When a limit involves expressions like (x + constant), substitution transforms it into a simpler form where the standard formulas apply directly.

Question 15. Evaluate \( \lim_{x \to 1} \left( \frac{x^n - 1}{x - 1} \right) \)
Answer: Using the general limit formula \( \lim_{x \to a} \frac{x^m - a^m}{x - a} = ma^{m-1} \), we apply it with the base 1 and exponent n. Therefore, \( \lim_{x \to 1} \frac{x^n - 1}{x - 1} = \lim_{x \to 1} \frac{x^n - 1^n}{x - 1} = n \cdot 1^{n - 1} = n \cdot 1 = n \).
In simple words: Apply the power limit formula: the exponent n comes down in front, and 1 raised to any power is 1, so the answer is just n.

Exam Tip: This result is a special case that appears frequently - when the limit point is 1, the formula simplifies dramatically because 1^k = 1 for any k.

Question 16. Evaluate \( \lim_{x \to a} \left( \frac{\sqrt{x} - \sqrt{a}}{x - a} \right) \)
Answer: Direct substitution yields \( \frac{0}{0} \), which is indeterminate. We rationalize the numerator by multiplying by the conjugate: \( \frac{\sqrt{x} - \sqrt{a}}{x - a} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}} = \frac{x - a}{(x - a)(\sqrt{x} + \sqrt{a})} = \frac{1}{\sqrt{x} + \sqrt{a}} \). Substituting: \( \lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a} = \lim_{x \to a} \frac{1}{\sqrt{x} + \sqrt{a}} = \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}} \).
In simple words: Multiply the fraction by the conjugate \(\sqrt{x} + \sqrt{a}\) over itself, simplify the indeterminate form, then substitute.

Exam Tip: Rationalization is a powerful technique for limits involving square roots - multiplying by the conjugate converts the difference of radicals into an algebraic expression.

Question 17. Evaluate \( \lim_{x \to a} \frac{\frac{1}{x^2} - \frac{1}{a^2}}{x - a} \)
Answer: To find this limit, we use L'Hospital's rule. Since both numerator and denominator approach zero as x approaches a, we differentiate each separately.

Taking the derivative of the numerator: \( \frac{d}{dx}\left(\frac{1}{x^2} - \frac{1}{a^2}\right) = -\frac{2}{x^3} \)

Taking the derivative of the denominator: \( \frac{d}{dx}(x - a) = 1 \)

By L'Hospital's rule:
\( \lim_{x \to a} \frac{\frac{1}{x^2} - \frac{1}{a^2}}{x - a} = \lim_{x \to a} \frac{-\frac{2}{x^3}}{1} = -\frac{2}{a^3} \)

We can also write this as \( \lim_{x \to a} \frac{\frac{1}{x^2} - \frac{1}{a^2}}{x - a} = \frac{1}{2\sqrt{a}} \) after careful algebraic simplification.
In simple words: When substituting the value directly gives 0/0, use L'Hospital's rule by taking derivatives of the top and bottom separately, then substitute again.

Exam Tip: Always check that the limit is indeterminate (0/0 or ∞/∞) before applying L'Hospital's rule; using it otherwise may lead to incorrect results.

Question 18. Evaluate \( \lim_{h \to 0} \frac{1}{h}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) \)
Answer: This limit has the indeterminate form 0/0 as h approaches 0. Using L'Hospital's rule, we differentiate numerator and denominator with respect to h.

The derivative of the numerator with respect to h is: \( \frac{d}{dh}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \frac{-1}{2\sqrt{x+h}} \)

The derivative of the denominator with respect to h is: 1

Therefore:
\( \lim_{h \to 0} \frac{1}{h}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \lim_{h \to 0} \frac{-\frac{1}{2\sqrt{x+h}}}{1} = \frac{-1}{2\sqrt{x}} \)

The value of this limit is \( \frac{-1}{2\sqrt{x}} \), which simplifies further to show the rate of change of \( \frac{1}{\sqrt{x}} \).
In simple words: When you have a fraction of fractions that gives 0/0, apply L'Hospital's rule to get a simpler result that you can evaluate.

Exam Tip: Confirm the indeterminate form exists before using L'Hospital's rule; this technique only works for 0/0 or ∞/∞ cases.

Question 19. Evaluate \( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \)
Answer: Direct substitution gives the indeterminate form 0/0. We apply L'Hospital's rule by differentiating numerator and denominator.

Differentiating the numerator: \( \frac{d}{dx}(\sqrt{1+x} - 1) = \frac{1}{2\sqrt{1+x}} \)

Differentiating the denominator: \( \frac{d}{dx}(x) = 1 \)

By L'Hospital's rule:
\( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}}}{1} = \frac{1}{2\sqrt{1}} = \frac{1}{2} \)

Therefore, the value of \( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \) is \( \frac{1}{2} \).
In simple words: This limit represents how fast the square root function grows near x = 0. Taking derivatives helps us find this rate of growth.

Exam Tip: For limits involving square roots, L'Hospital's rule is often faster than rationalizing; always look for the indeterminate form first.

Question 20. Evaluate \( \lim_{x \to 0} \frac{\sqrt{2-x} - \sqrt{2+x}}{x} \)
Answer: Substituting x = 0 directly yields 0/0, an indeterminate form. We use L'Hospital's rule to resolve this.

Differentiating the numerator: \( \frac{d}{dx}(\sqrt{2-x} - \sqrt{2+x}) = \frac{-1}{2\sqrt{2-x}} - \frac{1}{2\sqrt{2+x}} \)

Differentiating the denominator: \( \frac{d}{dx}(x) = 1 \)

Applying L'Hospital's rule:
\( \lim_{x \to 0} \frac{\sqrt{2-x} - \sqrt{2+x}}{x} = \lim_{x \to 0} \left(\frac{-1}{2\sqrt{2-x}} - \frac{1}{2\sqrt{2+x}}\right) = \frac{-1}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{-2}{2\sqrt{2}} = \frac{-1}{\sqrt{2}} \)

Hence, \( \lim_{x \to 0} \frac{\sqrt{2-x} - \sqrt{2+x}}{x} = \frac{-1}{\sqrt{2}} \).
In simple words: When a limit involves multiple square roots creating an indeterminate form, differentiate each term separately and add the results.

Exam Tip: Be careful with signs when differentiating; a negative inside the square root produces a negative derivative coefficient.

Question 21. Evaluate \( \lim_{x \to 0} \frac{\sqrt{1+x+x^2} - 1}{x} \)
Answer: Direct substitution gives 0/0, so we use L'Hospital's rule.

Differentiating the numerator: \( \frac{d}{dx}(\sqrt{1+x+x^2} - 1) = \frac{1+2x}{2\sqrt{1+x+x^2}} \)

Differentiating the denominator: \( \frac{d}{dx}(x) = 1 \)

By L'Hospital's rule:
\( \lim_{x \to 0} \frac{\sqrt{1+x+x^2} - 1}{x} = \lim_{x \to 0} \frac{1+2x}{2\sqrt{1+x+x^2}} = \frac{1+0}{2\sqrt{1+0+0}} = \frac{1}{2} \)

Therefore, the value of the limit is \( \frac{1}{2} \).
In simple words: Even when the expression under the square root is more complex, differentiating it using the chain rule and applying L'Hospital's rule gives the answer quickly.

Exam Tip: Use the chain rule correctly when differentiating composite functions like square roots of polynomials.

Question 22. Evaluate \( \lim_{x \to 0} \frac{\sqrt{3-x} - 1}{2-x} \)
Answer: When we substitute x = 0 directly, we get \( \frac{\sqrt{3} - 1}{2} \), which is not indeterminate. Using the continuity formula, we can evaluate this limit by direct substitution.

Since the function is continuous at x = 0:
\( \lim_{x \to 0} \frac{\sqrt{3-x} - 1}{2-x} = \frac{\sqrt{3-0} - 1}{2-0} = \frac{\sqrt{3} - 1}{2} \)

Therefore, the limit equals \( \frac{\sqrt{3} - 1}{2} \).
In simple words: Sometimes a limit can be found simply by plugging in the value of x directly, without needing L'Hospital's rule or other advanced techniques.

Exam Tip: Always check if direct substitution works before applying L'Hospital's rule - it saves time and reduces errors.

Question 23. Evaluate \( \lim_{x \to 0} \frac{2x}{\sqrt{a+x} - \sqrt{a-x}} \)
Answer: To evaluate this limit, we rationalize the denominator by multiplying both numerator and denominator by \( \sqrt{a+x} + \sqrt{a-x} \).

Multiplying numerator and denominator by \( \sqrt{a+x} + \sqrt{a-x} \):
\( \lim_{x \to 0} \frac{2x}{\sqrt{a+x} - \sqrt{a-x}} = \lim_{x \to 0} \frac{2x(\sqrt{a+x} + \sqrt{a-x})}{(\sqrt{a+x} - \sqrt{a-x})(\sqrt{a+x} + \sqrt{a-x})} \)

The denominator simplifies using the difference of squares:
\( (\sqrt{a+x})^2 - (\sqrt{a-x})^2 = (a+x) - (a-x) = 2x \)

Therefore:
\( \lim_{x \to 0} \frac{2x}{\sqrt{a+x} - \sqrt{a-x}} = \lim_{x \to 0} \frac{2x(\sqrt{a+x} + \sqrt{a-x})}{2x} = \lim_{x \to 0} (\sqrt{a+x} + \sqrt{a-x}) = 2\sqrt{a} \)

The limit equals \( 2\sqrt{a} \).
In simple words: Rationalize by multiplying by the conjugate - the denominator becomes a simple expression and the x terms cancel out.

Exam Tip: Rationalization is often quicker than L'Hospital's rule for limits involving conjugate square root expressions.

Question 24. Evaluate \( \lim_{x \to 1} \frac{\sqrt{3+x} - \sqrt{5-x}}{x^2 - 1} \)
Answer: Substituting x = 1 gives 0/0, so this is indeterminate. We use L'Hospital's rule.

Differentiating the numerator: \( \frac{d}{dx}(\sqrt{3+x} - \sqrt{5-x}) = \frac{1}{2\sqrt{3+x}} + \frac{1}{2\sqrt{5-x}} \)

Differentiating the denominator: \( \frac{d}{dx}(x^2 - 1) = 2x \)

Applying L'Hospital's rule:
\( \lim_{x \to 1} \frac{\sqrt{3+x} - \sqrt{5-x}}{x^2 - 1} = \lim_{x \to 1} \frac{\frac{1}{2\sqrt{3+x}} + \frac{1}{2\sqrt{5-x}}}{2x} \)

Evaluating at x = 1:
\( = \frac{\frac{1}{2\sqrt{4}} + \frac{1}{2\sqrt{4}}}{2(1)} = \frac{\frac{1}{4} + \frac{1}{4}}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \)

The limit equals \( \frac{1}{4} \).
In simple words: When L'Hospital's rule applies, differentiate the top and bottom separately, then substitute the limiting value into both derivatives.

Exam Tip: Double-check that the answer is no longer in indeterminate form after applying L'Hospital's rule - if it still is, apply the rule again.

Question 25. Evaluate \( \lim_{x \to 2} \frac{x^2 - 4}{\sqrt{x+2} - \sqrt{3x-2}} \)
Answer: At x = 2, we get 0/0, so we apply L'Hospital's rule to find this limit.

Differentiating the numerator: \( \frac{d}{dx}(x^2 - 4) = 2x \)

Differentiating the denominator: \( \frac{d}{dx}(\sqrt{x+2} - \sqrt{3x-2}) = \frac{1}{2\sqrt{x+2}} - \frac{3}{2\sqrt{3x-2}} \)

By L'Hospital's rule:
\( \lim_{x \to 2} \frac{x^2 - 4}{\sqrt{x+2} - \sqrt{3x-2}} = \lim_{x \to 2} \frac{2x}{\frac{1}{2\sqrt{x+2}} - \frac{3}{2\sqrt{3x-2}}} \)

At x = 2:
\( = \frac{2(2)}{\frac{1}{2\sqrt{4}} - \frac{3}{2\sqrt{4}}} = \frac{4}{\frac{1}{4} - \frac{3}{4}} = \frac{4}{\frac{-2}{4}} = \frac{4}{-\frac{1}{2}} = -8 \)

The value of the limit is -8.
In simple words: Apply L'Hospital's rule, being careful with the derivative of each term, especially negative terms in the denominator.

Exam Tip: When differentiating square roots in denominators, remember the chain rule produces a coefficient in the numerator of the resulting fraction.

Question 26. Evaluate \( \lim_{x \to 4} \frac{3 - \sqrt{5+x}}{1 - \sqrt{5-x}} \)
Answer: To evaluate this limit, we rationalize by multiplying numerator and denominator by their respective conjugates: \( (1 + \sqrt{5-x})(3 + \sqrt{5+x}) \).

Multiplying by the conjugate pairs:
\( \lim_{x \to 4} \frac{3 - \sqrt{5+x}}{1 - \sqrt{5-x}} = \lim_{x \to 4} \frac{(3 - \sqrt{5+x})(1 + \sqrt{5-x})(3 + \sqrt{5+x})}{(1 - \sqrt{5-x})(1 + \sqrt{5-x})(3 + \sqrt{5+x})} \)

The numerator \( 3 - \sqrt{5+x} \) times its conjugate gives: \( (3)^2 - (5+x) = 9 - 5 - x = 4 - x \)

The denominator \( 1 - \sqrt{5-x} \) times its conjugate gives: \( (1)^2 - (5-x) = 1 - 5 + x = x - 4 \)

Therefore:
\( \lim_{x \to 4} \frac{3 - \sqrt{5+x}}{1 - \sqrt{5-x}} = \lim_{x \to 4} \frac{(4-x)(1 + \sqrt{5-x})(3 + \sqrt{5+x})}{(x-4)(3 + \sqrt{5+x})} \)

Since \( 4 - x = -(x-4) \):
\( = \lim_{x \to 4} \frac{-(1 + \sqrt{5-x})(3 + \sqrt{5+x})}{(3 + \sqrt{5+x})} = \lim_{x \to 4} -(1 + \sqrt{5-x}) = -(1 + 1) = -\frac{1}{3} \)

The limit equals \( -\frac{1}{3} \).
In simple words: When both numerator and denominator contain square roots, rationalize both parts separately and look for common factors that cancel.

Exam Tip: Keep track of negative signs when dealing with conjugates - a factor of (a - b) in one part might appear as -(b - a) elsewhere, allowing cancellation.

Question 27. Evaluate \( \lim_{x \to 0} \frac{\sqrt{a+x} - \sqrt{a}}{x\sqrt{a(a+x)}} \)
Answer: At x = 0, both numerator and denominator equal zero, so we apply L'Hospital's rule.

Differentiating the numerator: \( \frac{d}{dx}(\sqrt{a+x} - \sqrt{a}) = \frac{1}{2\sqrt{a+x}} \)

Differentiating the denominator: \( \frac{d}{dx}(x\sqrt{a(a+x)}) = \sqrt{a(a+x)} + x \cdot \sqrt{a} \cdot \frac{1}{2\sqrt{a+x}} = \sqrt{a(a+x)} + \frac{ax}{2\sqrt{a(a+x)}} \)

At x = 0:
\( = \frac{\frac{1}{2\sqrt{a}}}{\sqrt{a \cdot a} + 0} = \frac{\frac{1}{2\sqrt{a}}}{a} = \frac{1}{2a\sqrt{a}} = \frac{1}{2a^{3/2}} \)

Alternatively, this can be expressed as \( \frac{1}{2a\sqrt{a}} \).
In simple words: After applying L'Hospital's rule, substitute the limiting value into the derivative expressions to get the final answer.

Exam Tip: For products in the denominator, use the product rule when differentiating - do not forget any terms.

Question 28. Evaluate \( \lim_{x \to 0} \frac{\sqrt{1+x^2} - \sqrt{1+x}}{\sqrt{1+x^3} - \sqrt{1+x}} \)
Answer: Substituting x = 0 gives 0/0, which is indeterminate. We use L'Hospital's rule.

Differentiating the numerator: \( \frac{d}{dx}(\sqrt{1+x^2} - \sqrt{1+x}) = \frac{2x}{2\sqrt{1+x^2}} - \frac{1}{2\sqrt{1+x}} = \frac{x}{\sqrt{1+x^2}} - \frac{1}{2\sqrt{1+x}} \)

Differentiating the denominator: \( \frac{d}{dx}(\sqrt{1+x^3} - \sqrt{1+x}) = \frac{3x^2}{2\sqrt{1+x^3}} - \frac{1}{2\sqrt{1+x}} \)

By L'Hospital's rule:
\( \lim_{x \to 0} \frac{\sqrt{1+x^2} - \sqrt{1+x}}{\sqrt{1+x^3} - \sqrt{1+x}} = \lim_{x \to 0} \frac{\frac{x}{\sqrt{1+x^2}} - \frac{1}{2\sqrt{1+x}}}{\frac{3x^2}{2\sqrt{1+x^3}} - \frac{1}{2\sqrt{1+x}}} \)

At x = 0:
\( = \frac{0 - \frac{1}{2}}{0 - \frac{1}{2}} = \frac{-\frac{1}{2}}{-\frac{1}{2}} = -1 \)

The value of the limit is -1.
In simple words: When both numerator and denominator have multiple terms, differentiate each term separately and recombine them before evaluating.

Exam Tip: After applying L'Hospital's rule and evaluating at the limiting value, verify your arithmetic carefully as sign errors are common.

Question 29. Evaluate \( \lim_{x \to 1} \frac{x^4 - 3x^2 + 2}{x^3 - 5x^2 + 3x + 1} \)
Answer: Substituting x = 1 into both numerator and denominator yields 0/0. We apply L'Hospital's rule.

Differentiating the numerator: \( \frac{d}{dx}(x^4 - 3x^2 + 2) = 4x^3 - 6x \)

Differentiating the denominator: \( \frac{d}{dx}(x^3 - 5x^2 + 3x + 1) = 3x^2 - 10x + 3 \)

By L'Hospital's rule:
\( \lim_{x \to 1} \frac{x^4 - 3x^2 + 2}{x^3 - 5x^2 + 3x + 1} = \lim_{x \to 1} \frac{4x^3 - 6x}{3x^2 - 10x + 3} \)

At x = 1:
\( = \frac{4(1)^3 - 6(1)}{3(1)^2 - 10(1) + 3} = \frac{4 - 6}{3 - 10 + 3} = \frac{-2}{-4} = \frac{1}{2} \)

Therefore, the limit equals \( \frac{1}{2} \).
In simple words: For polynomial limits that are indeterminate, L'Hospital's rule involves taking the derivative of each polynomial and then substituting.

Exam Tip: Make sure to apply the power rule correctly when differentiating polynomial terms - the coefficient gets multiplied by the original exponent.

Question 30. Evaluate \( \lim_{x \to 2} \frac{3^x - 3^{3-x} - 12}{3^{3-x} - 3^{x/2}} \)
Answer: At x = 2, substituting gives 0/0, which is indeterminate. We use L'Hospital's rule to evaluate this exponential limit.

Differentiating the numerator: \( \frac{d}{dx}(3^x - 3^{3-x} - 12) = 3^x \ln 3 - 3^{3-x} \cdot (-\ln 3) = 3^x \ln 3 + 3^{3-x} \ln 3 \)

Differentiating the denominator: \( \frac{d}{dx}(3^{3-x} - 3^{x/2}) = 3^{3-x} \cdot (-\ln 3) - 3^{x/2} \cdot \frac{\ln 3}{2} = -3^{3-x} \ln 3 - \frac{3^{x/2} \ln 3}{2} \)

By L'Hospital's rule:
\( \lim_{x \to 2} \frac{3^x - 3^{3-x} - 12}{3^{3-x} - 3^{x/2}} = \lim_{x \to 2} \frac{3^x \ln 3 + 3^{3-x} \ln 3}{-3^{3-x} \ln 3 - \frac{3^{x/2} \ln 3}{2}} \)

At x = 2:
\( = \frac{3^2 \ln 3 + 3^1 \ln 3}{-3^1 \ln 3 - \frac{3 \ln 3}{2}} = \frac{(9 + 3) \ln 3}{(-3 - \frac{3}{2}) \ln 3} = \frac{12}{-\frac{9}{2}} = \frac{12 \cdot 2}{-9} = -\frac{8}{3} \)

The limit equals \( -\frac{8}{3} \).
In simple words: For exponential limits, remember that the derivative of \( a^x \) is \( a^x \ln a \), and use the chain rule for exponents that depend on x.

Exam Tip: Keep the natural logarithm factors consistent throughout differentiation - they factor out at the end and often cancel between numerator and denominator.

Question 31. Evaluate \( \lim_{x \to 0} \left( \frac{e^{4x} - 1}{x} \right) \)
Answer: As \( x \to 0 \), we obtain \( \frac{e^{4x} - 1}{x} = \frac{0}{0} \), which is an indeterminate form. Using L'Hospital's rule, we differentiate the numerator and denominator:
\( \lim_{x \to 0} \frac{e^{4x} - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^{4x} - 1)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{4e^{4x}}{1} = \lim_{x \to 0} 4e^{4x} = 4 \)
In simple words: When you substitute 0 into the original expression, both the top and bottom become 0. Apply L'Hospital's rule by taking derivatives of the numerator and denominator separately, then evaluate the limit. The answer is 4.

Exam Tip: Always confirm that direct substitution gives the indeterminate form 0/0 or ∞/∞ before applying L'Hospital's rule. This rule is your key tool for resolving these forms.

Question 32. Evaluate \( \lim_{x \to 0} \left( \frac{e^{2+x} - e^2}{x} \right) \)
Answer: As \( x \to 0 \), we obtain \( \frac{e^{2+x} - e^2}{x} = \frac{0}{0} \), which is an indeterminate form. Applying L'Hospital's rule, we differentiate numerator and denominator:
\( \lim_{x \to 0} \frac{e^{2+x} - e^2}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^{2+x} - e^2)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{e^{2+x}}{1} = e^2 \)
In simple words: Direct substitution gives 0/0, which is indeterminate. Use L'Hospital's rule to find the derivative of the top and bottom separately. The limit evaluates to e squared.

Exam Tip: Recognize that \( e^{2+x} \) differentiates to itself (since the exponent has a constant part). This pattern holds for any exponential function.

Question 33. Evaluate \( \lim_{x \to 4} \left( \frac{e^x - e^4}{x - 4} \right) \)
Answer: As \( x \to 4 \), we obtain \( \frac{e^4 - e^4}{4 - 4} = \frac{0}{0} \), which is an indeterminate form. Using L'Hospital's rule, we differentiate numerator and denominator:
\( \lim_{x \to 4} \frac{e^x - e^4}{x - 4} = \lim_{x \to 4} \frac{\frac{d}{dx}(e^x - e^4)}{\frac{d}{dx}(x - 4)} = \lim_{x \to 4} \frac{e^x}{1} = e^4 \)
In simple words: Plugging in x = 4 makes both numerator and denominator equal to zero. Apply L'Hospital's rule by differentiating the top and bottom. You get e to the power of 4.

Exam Tip: This example demonstrates that \( \lim_{x \to a} \frac{e^x - e^a}{x - a} = e^a \), a useful derivative-definition pattern to remember.

Question 34. Evaluate \( \lim_{x \to 0} \left( \frac{e^{3x} - e^{2x}}{x} \right) \)
Answer: As \( x \to 0 \), we obtain \( \frac{e^0 - e^0}{0} = \frac{0}{0} \), which is an indeterminate form. Applying L'Hospital's rule:
\( \lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^{3x} - e^{2x})}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{3e^{3x} - 2e^{2x}}{1} = 3 - 2 = 1 \)
In simple words: Direct substitution gives 0/0. Differentiate the numerator to get 3 times e to the 3x minus 2 times e to the 2x, and the denominator becomes 1. Substitute x = 0 and compute: 3 - 2 = 1.

Exam Tip: When the numerator contains a difference of exponentials with different exponents, apply the chain rule carefully - each exponent contributes its coefficient when you differentiate.

Question 35. Evaluate \( \lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x} \)
Answer: As \( x \to 0 \), we get \( \frac{0}{0} \), an indeterminate form. Applying L'Hospital's rule, differentiate the top and bottom:
\( \lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x} = \lim_{x \to 0} \frac{3e^{3x} - 2e^{2x}}{1} = 3(1) - 2(1) = 1 \)
In simple words: The top and bottom are both zero when x equals 0. Use L'Hospital's rule: differentiate the numerator to get 3e to the 3x minus 2e to the 2x, and the denominator becomes 1. At x = 0, this gives 3 minus 2, which equals 1.

Exam Tip: Always apply the chain rule when differentiating exponential functions with coefficients in the exponent - do not forget the multiplier in front.

Question 36. Evaluate \( \lim_{x \to 0} \left( \frac{e^{bx} - e^{ax}}{x} \right), 0 < a < b \)
Answer: As \( x \to 0 \), we obtain \( \frac{e^0 - e^0}{0} = \frac{0}{0} \), an indeterminate form. Using L'Hospital's rule:
\( \lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} = \lim_{x \to 0} \frac{be^{bx} - ae^{ax}}{1} = b(1) - a(1) = b - a \)
In simple words: When you plug in x = 0, you get 0/0. Apply L'Hospital's rule by differentiating: the top becomes b times e to the bx minus a times e to the ax, and the bottom becomes 1. Setting x = 0 gives b minus a.

Exam Tip: This result shows that the limit of a difference of exponentials with different coefficients is simply the difference of those coefficients - a handy formula to recognize quickly.

Question 37. Evaluate \( \lim_{x \to 0} \left( \frac{a^x - b^x}{x} \right) \)
Answer: As \( x \to 0 \), we obtain \( \frac{a^0 - b^0}{0} = \frac{1 - 1}{0} = \frac{0}{0} \), an indeterminate form. Applying L'Hospital's rule:
\( \lim_{x \to 0} \frac{a^x - b^x}{x} = \lim_{x \to 0} \frac{a^x \ln a - b^x \ln b}{1} = \ln a - \ln b = \ln \frac{a}{b} \)
In simple words: Direct substitution gives 0/0. Use L'Hospital's rule: the derivative of a to the x is a to the x times the natural log of a, and similarly for b to the x. At x = 0, you get ln(a) minus ln(b), which can be written as the natural log of a divided by b.

Exam Tip: Remember that the derivative of \( a^x \) is \( a^x \ln a \). This involves the logarithm of the base, not the exponent.

Question 38. Evaluate \( \lim_{x \to 0} \left( \frac{a^x - a^{-x}}{x} \right) \)
Answer: As \( x \to 0 \), we have \( \frac{a^0 - a^0}{0} = \frac{0}{0} \), an indeterminate form. Applying L'Hospital's rule:
\( \lim_{x \to 0} \frac{a^x - a^{-x}}{x} = \lim_{x \to 0} \frac{a^x \ln a - (-a^{-x} \ln a)}{1} = \lim_{x \to 0} \frac{a^x \ln a + a^{-x} \ln a}{1} = \ln a + \ln a = 2 \ln a \)
In simple words: Substituting x = 0 gives 0/0. Differentiate using L'Hospital's rule: the derivative of a to the x is a to the x times ln(a), and the derivative of a to the negative x is negative a to the negative x times ln(a). At x = 0, both parts equal ln(a), giving a total of 2 ln(a).

Exam Tip: Pay careful attention to the chain rule when differentiating \( a^{-x} \): the negative sign from the exponent contributes an extra negative that cancels the negative derivative.

Question 39. Evaluate \( \lim_{x \to 0} \left( \frac{2^x - 1}{x} \right) \)
Answer: As \( x \to 0 \), we obtain \( \frac{2^0 - 1}{0} = \frac{0}{0} \), an indeterminate form. Using L'Hospital's rule:
\( \lim_{x \to 0} \frac{2^x - 1}{x} = \lim_{x \to 0} \frac{2^x \ln 2}{1} = \ln 2 \)
In simple words: Plugging in x = 0 gives 0/0. Apply L'Hospital's rule: differentiate 2 to the x to get 2 to the x times ln(2), and the denominator becomes 1. At x = 0, this equals ln(2).

Exam Tip: This is the definition of the derivative of \( 2^x \) at x = 0. For any base a, \( \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \).

Question 40. Evaluate \( \lim_{x \to 0} \left( \frac{3^{2+x} - 9}{x} \right) \)
Answer: As \( x \to 0 \), we have \( \frac{3^2 - 9}{0} = \frac{0}{0} \), an indeterminate form. Applying L'Hospital's rule:
\( \lim_{x \to 0} \frac{3^{2+x} - 9}{x} = \lim_{x \to 0} \frac{3^{2+x} \ln 3}{1} = 3^2 \ln 3 = 9 \ln 3 \)
In simple words: Direct substitution gives 0/0. Use L'Hospital's rule: the derivative of 3 to the (2+x) is 3 to the (2+x) times ln(3), and the denominator becomes 1. At x = 0, you get 3 squared (which is 9) times ln(3), giving 9 ln(3).

Exam Tip: When the exponent contains a constant plus a variable, the derivative gives the original function times the natural logarithm of the base - the constant does not affect the result.

Question 41. Evaluate the following limits: \( \lim_{x \to 0} \frac{\sin 4x}{6x} \)
Answer: As \( x \to 0 \), we get \( \frac{0}{0} \), an indeterminate form. Using the standard limit formula \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\( \lim_{x \to 0} \frac{\sin 4x}{6x} = \lim_{x \to 0} \left( \frac{\sin 4x}{4x} \times \frac{4}{6} \right) = 1 \times \frac{2}{3} = \frac{2}{3} \)
In simple words: When x approaches 0, both the numerator and denominator approach 0. Rewrite the expression by extracting the coefficient 4 from inside the sine function, then use the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Multiply by the remaining coefficient to get 2/3.

Exam Tip: Always look for the pattern \( \frac{\sin(kx)}{kx} \) inside complex expressions - this is the fastest route to the answer without needing L'Hospital's rule.

Question 42. Evaluate the following limits: \( \lim_{x \to 0} \frac{\sin 5x}{\sin 8x} \)
Answer: As \( x \to 0 \), we obtain \( \frac{0}{0} \), an indeterminate form. Using the standard limit formula \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\( \lim_{x \to 0} \frac{\sin 5x}{\sin 8x} = \lim_{x \to 0} \left( \frac{\sin 5x}{5x} \times \frac{8x}{\sin 8x} \times \frac{5x}{8x} \right) = 1 \times 1 \times \frac{5}{8} = \frac{5}{8} \)
In simple words: Direct substitution gives 0/0. Rewrite the expression by inserting the coefficients 5x in the numerator and 8x in the denominator. Each fraction \( \frac{\sin kx}{kx} \) approaches 1 as x approaches 0. The remaining ratio of coefficients is 5/8.

Exam Tip: For limits of the form \( \frac{\sin(ax)}{\sin(bx)} \), the answer is always \( \frac{a}{b} \) as x approaches 0 - recognize this pattern to save time.

Question 43. Evaluate the following limits: \( \lim_{x \to 0} \frac{\tan 3x}{\tan 5x} \)
Answer: As \( x \to 0 \), we have \( \frac{0}{0} \), an indeterminate form. Using the standard limit formula \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):
\( \lim_{x \to 0} \frac{\tan 3x}{\tan 5x} = \lim_{x \to 0} \left( \frac{\tan 3x}{3x} \times \frac{5x}{\tan 5x} \times \frac{3x}{5x} \right) = 1 \times 1 \times \frac{3}{5} = \frac{3}{5} \)
In simple words: Substituting x = 0 gives 0/0. Restructure by introducing 3x in the numerator and 5x in the denominator. Each ratio \( \frac{\tan kx}{kx} \) equals 1 as x approaches 0. The coefficient ratio is 3/5.

Exam Tip: For \( \lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} \), the limit is simply \( \frac{a}{b} \) - this is a useful shortcut that applies whenever both the numerator and denominator involve only tangent functions with linear arguments.

Question 44. Evaluate the following limits: \( \lim_{x \to 0} \frac{\tan \alpha x}{\tan \beta x} \)
Answer: As \( x \to 0 \), we obtain \( \frac{0}{0} \), an indeterminate form. Using the standard limit \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):
\( \lim_{x \to 0} \frac{\tan \alpha x}{\tan \beta x} = \lim_{x \to 0} \left( \frac{\tan \alpha x}{\alpha x} \times \frac{\beta x}{\tan \beta x} \times \frac{\alpha x}{\beta x} \right) = 1 \times 1 \times \frac{\alpha}{\beta} = \frac{\alpha}{\beta} \)
In simple words: Plugging in x = 0 gives 0/0. Insert \( \alpha x \) in the numerator and \( \beta x \) in the denominator to expose the standard limit pattern. Each term \( \frac{\tan kx}{kx} \) approaches 1. The ratio of coefficients gives \( \frac{\alpha}{\beta} \).

Exam Tip: This general form confirms that \( \lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} = \frac{a}{b} \) for any constants a and b - memorizing this saves significant calculation time on exams.

Question 5. Evaluate the following limits:
Answer: To find the limit, first verify the form of the limit. Apply this method if the limit satisfies any one of the seven indeterminate forms.
In simple words: Check which indeterminate form the limit matches before selecting a solving method.

Exam Tip: Always identify the indeterminate form (0/0, ∞/∞, 0×∞, etc.) before attempting to evaluate the limit.

Question 6. Evaluate the following limits:
Answer: To find the limit, first verify the form of the limit. Apply this method if the limit satisfies any one of the seven indeterminate forms.
In simple words: Check which indeterminate form the limit matches before selecting a solving method.

Exam Tip: Always identify the indeterminate form (0/0, ∞/∞, 0×∞, etc.) before attempting to evaluate the limit.

Question 6. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin 4x}{\tan 7x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

So \( \lim_{x \to 0} \frac{\sin 4x}{\tan 7x} = \lim_{x \to 0} \left( \frac{\sin 4x}{4x} \right) \times \frac{4x}{\sin 7x} \times \frac{7x}{4x} = \frac{4}{7} \)

Therefore, \( \lim_{x \to 0} \frac{\sin 4x}{\tan 7x} = \frac{4}{7} \)
In simple words: Rewrite the expression using standard limit formulas, then simplify to get the answer.

Exam Tip: Remember the key formulas \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \) - they are the foundation for solving trigonometric limit problems.

Question 7. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\tan 3x}{\sin 4x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \) and \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

So \( \lim_{x \to 0} \frac{\tan 3x}{\sin 4x} = \lim_{x \to 0} \left( \frac{\tan 3x}{3x} \right) \times \frac{3x}{\sin 4x} \times \frac{4x}{4x} = \frac{3}{4} \)

Therefore, \( \lim_{x \to 0} \frac{\tan 3x}{\sin 4x} = \frac{3}{4} \)
In simple words: Apply the standard formulas and rearrange to isolate each variable's coefficient.

Exam Tip: When both sine and tangent appear, use their respective limit definitions separately and multiply the results.

Question 8. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin x - 2 \sin 3x + \sin 5x}{x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

So \( \lim_{x \to 0} \frac{\sin x - 2 \sin 3x + \sin 5x}{x} = \lim_{x \to 0} \left( \frac{\sin x}{x} - \frac{2\sin 3x}{x} + \frac{\sin 5x}{x} \right) = \lim_{x \to 0} \left( \frac{\sin x}{x} - \frac{2\sin 3x}{3x} \times 3 + \frac{\sin 5x}{5x} \times 5 \right) \)

Using the formula above, we get: \( 1 - 2 \times 3 + 5 = 0 \)

Therefore, \( \lim_{x \to 0} \frac{\sin x - 2 \sin 3x + \sin 5x}{x} = 0 \)
In simple words: Separate the terms and apply the standard sine formula to each one, then add the results together.

Exam Tip: Break down complex expressions into simpler fractions that match standard limit formulas.

Question 9. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{6}} \frac{(2 \sin^2 x + \sin x - 1)}{(2 \sin^2 x - 3\sin x + 1)} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).
In simple words: Substitute the limit value to check if the expression is indeterminate, then apply appropriate techniques to resolve it.

Exam Tip: When direct substitution gives an indeterminate form, try factoring the numerator and denominator to cancel common factors.

Question 10. Evaluate the following limits:
\( \lim_{x \to 0} \frac{(\sin 2x + 3x)}{(2x + \sin 3x)} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) or L'Hôpital's Rule can be applied.

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{\sin 2x + 3x}{2x + \sin 3x} = \lim_{x \to 0} \frac{\frac{\sin 2x}{x} + 3}{\frac{2x}{\sin 3x} + 1} = \lim_{x \to 0} \frac{\frac{\sin 2x}{x} + 3}{2 + \frac{\sin 3x}{x}} = \lim_{x \to 0} \frac{\frac{2\sin 2x}{2x} + 3}{2 + \frac{\sin 3x}{3x} \times 3} = \frac{2 + 3}{2 + 3} = 1 \)

Alternatively, by using L'Hôpital's Rule: Differentiate numerator and denominator

\( \lim_{x \to 0} \frac{2\cos 2x + 3}{2 + 3\cos 3x} = \frac{2 + 3}{2 + 3} = 1 \)

Therefore, \( \lim_{x \to 0} \frac{\sin 2x + 3x}{2x + \sin 3x} = 1 \)
In simple words: Divide both numerator and denominator by x, then apply standard formulas, or use differentiation when needed.

Exam Tip: L'Hôpital's Rule is a powerful alternative when direct algebraic manipulation becomes complex.

Question 11. Evaluate the following limits:
\( \lim_{x \to 0} \frac{(\tan 2x - x)}{(3x - \tan x)} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \) or L'Hôpital's Rule can be applied.

So, using the above formula, we have:

Divide numerator and denominator by x,

\( \lim_{x \to 0} \frac{\tan 2x - x}{3x - \tan x} = \lim_{x \to 0} \frac{\frac{\tan 2x}{x} - 1}{3 - \frac{\tan x}{x}} = \lim_{x \to 0} \frac{\frac{\tan 2x}{2x} \times 2 - 1}{3 - \frac{\tan x}{x}} = \lim_{x \to 0} \frac{2 \times \frac{\tan 2x}{2x} - 1}{3 - \frac{\tan x}{x}} = \frac{2 - 1}{3 - 1} = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} \frac{\tan 2x - x}{3x - \tan x} = \frac{1}{2} \)
In simple words: Divide by x to extract coefficients, apply standard tangent formulas, then compute the result.

Exam Tip: Watch the coefficients carefully when you have expressions like \( \tan(kx) \) - they scale the standard formula result.

Question 12. Evaluate the following limits:
\( \lim_{x \to 0} \frac{(x^2 - \tan 2x)}{\tan x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \) or L'Hôpital's Rule can be applied.

So, using the above formula, we have:

Divide numerator and denominator by x,

\( \lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x} = \lim_{x \to 0} \frac{\frac{x^2}{x} - \frac{\tan 2x}{x}}{\frac{\tan x}{x}} = \lim_{x \to 0} \frac{x - \frac{\tan 2x}{2x} \times 2}{1} = \lim_{x \to 0} \frac{x - 2 \times \frac{\tan 2x}{2x}}{1} = \frac{0 - 2}{1} = -2 \)

Therefore, \( \lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x} = -2 \)
In simple words: Divide by x in both parts, use the tangent formula, and simplify step by step.

Exam Tip: When x appears as a separate term, include it in your division to properly handle all parts of the expression.

Question 13. Evaluate the following limits:
\( \lim_{x \to 0} \frac{x\cos x + \sin x}{x^2 + \tan x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

So, using the above formula, we have:

Divide numerator and denominator by x,

\( \lim_{x \to 0} \frac{x\cos x + \sin x}{x^2 + \tan x} = \lim_{x \to 0} \frac{\frac{x\cos x}{x} + \frac{\sin x}{x}}{\frac{x^2}{x} + \frac{\tan x}{x}} = \lim_{x \to 0} \frac{\cos x + \frac{\sin x}{x}}{x + \frac{\tan x}{x}} = \frac{1 + 1}{0 + 1} = 2 \)

Therefore, \( \lim_{x \to 0} \frac{x\cos x + \sin x}{x^2 + \tan x} = 2 \)
In simple words: Split the fraction into separate terms and apply standard sine and tangent formulas to each.

Exam Tip: Always verify which terms vanish as x approaches 0 to simplify your calculation.

Question 14. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\tan x - \sin x}{\sin^2 x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Note: \( \tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \frac{\sin x - \sin x \cos x}{\cos x} = \sin x \left( \frac{1 - \cos x}{\cos x} \right) \)

\( \lim_{x \to 0} \frac{\tan x - \sin x}{\sin^2 x} = \lim_{x \to 0} \frac{\frac{1 - \cos x}{\cos x}}{\sin^2 x} = \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x \cos x} \)

Divide numerator and denominator by \( x^2 \),

\( = \lim_{x \to 0} \frac{\frac{1 - \cos x}{x^2}}{\frac{\sin^2 x \cos x}{x^2}} \)

Formula used: \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) and \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) or L'Hôpital's Rule can be used.

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{\frac{1 - \cos x}{x^2}}{\frac{\sin^2 x \cos x}{x^2}} = \frac{\frac{1}{2}}{1 \times 1} = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} \frac{\tan x - \sin x}{\sin^2 x} = \frac{1}{2} \)
In simple words: Rewrite tangent in terms of sine and cosine, simplify the resulting fraction, then use standard limit formulas.

Exam Tip: The key formula \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) often appears in trigonometric limits - memorize it.

Question 15. Evaluate the following limits:
\( \lim_{x \to 0} x \cosec x \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( 0 \times \infty \).

Formula used: \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \)

So, using the above formula, we have:

\( \lim_{x \to 0} x \cosec x = \lim_{x \to 0} \frac{x}{\sin x} = 1 \)

Therefore, \( \lim_{x \to 0} x \cosec x = 1 \)
In simple words: Rewrite cosecant as the reciprocal of sine, then apply the standard formula.

Exam Tip: Converting trigonometric functions to their basic forms (sine and cosine) often simplifies the problem.

Question 16. Evaluate the following limits:
\( \lim_{x \to 0} (x \cot 2x) \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( 0 \times \infty \).

Formula used: \( \lim_{x \to 0} \frac{x}{\tan x} = 1 \)

So, using the above formula, we have:

\( \lim_{x \to 0} x \cot 2x = \lim_{x \to 0} \frac{2x}{2\tan 2x} = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} x \cot 2x = \frac{1}{2} \)
In simple words: Express cotangent as the reciprocal of tangent, manipulate to match the standard formula, and solve.

Exam Tip: When coefficients appear inside trigonometric functions, extract them carefully and adjust the fraction accordingly.

Question 17. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin x \cos x}{3x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{\sin x \cos x}{3x} = \lim_{x \to 0} \frac{\sin x}{x} \times \frac{\cos x}{3} = 1 \times \frac{1}{3} = \frac{1}{3} \)

Therefore, \( \lim_{x \to 0} \frac{\sin x \cos x}{3x} = \frac{1}{3} \)
In simple words: Extract the sine term to match the standard formula, then evaluate the remaining factors.

Exam Tip: Separate the sine portion from other terms to apply the \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) formula effectively.

Question 18. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin(x / 4)}{x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{\sin(x/4)}{x} = \lim_{x \to 0} \frac{\sin(x/4)}{4(x/4)} = \frac{1}{4} \)

Therefore, \( \lim_{x \to 0} \frac{\sin(x/4)}{x} = \frac{1}{4} \)
In simple words: Rewrite the denominator to match the argument inside the sine function, then use the standard limit.

Exam Tip: The key is to make the denominator match the argument of the sine function to apply the standard formula.

Question 19. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\tan(x / 2)}{3x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{\tan(x/2)}{3x} = \lim_{x \to 0} \frac{\tan(x/2)}{6(x/2)} = \frac{1}{6} \) [Divide and multiply by 2 on denominator]

Therefore, \( \lim_{x \to 0} \frac{\tan(x/2)}{3x} = \frac{1}{6} \)
In simple words: Adjust the denominator to match the tangent argument, apply the standard formula, and simplify.

Exam Tip: When fractions appear inside trigonometric arguments, manipulate the denominator to match before applying formulas.

Question 20. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Note: \( 1 - \cos x = 2 \sin^2(x/2) \)

Formula used: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x} = \lim_{x \to 0} \frac{2 \sin^2(x/2)}{\sin^2 x} \)

Divide numerator and denominator by \( x^2 \), we have:

\( \lim_{x \to 0} \frac{2 \sin^2(x/2)}{x^2} = \lim_{x \to 0} \frac{2 \sin^2(x/2)}{x^2} = \lim_{x \to 0} \frac{2 \left[ \frac{\sin(x/2)}{x/2} \right]^2 \times (x/2)^2}{x^2} = \lim_{x \to 0} \frac{2 \times 1 \times (x^2/4)}{x^2} = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x} = \frac{1}{2} \)
In simple words: Use the identity for 1 - cos x, divide by x squared, and apply the sine formula to each part.

Exam Tip: The identity \( 1 - \cos x = 2 \sin^2(x/2) \) is essential for simplifying many trigonometric limits.

Question 21. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \)

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{9[1 - \cos 3x]}{(3x)^2} = \frac{9}{2} \)

Therefore, \( \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \frac{9}{2} \)
In simple words: Apply the standard formula for 1 - cos, accounting for the coefficient inside the cosine function.

Exam Tip: When a coefficient k appears inside \( \cos(kx) \), the limit becomes \( \frac{k^2}{2} \) - remember this pattern.

Question 22. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 2x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) and \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

Divide numerator and denominator by \( x^2 \), we have:

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 2x} = \lim_{x \to 0} \frac{\frac{[1 - \cos x]}{x^2}}{\frac{\sin^2 2x}{x^2}} = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 2x} = \frac{1}{2} \)
In simple words: Divide both parts by x squared, apply the standard formulas, and evaluate the result.

Exam Tip: Dividing by \( x^2 \) is a powerful technique for handling expressions with 1 - cos and squared trigonometric functions.

Question 23. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos 2x}{3\tan^2 x} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

Formula used: \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) and \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

Divide numerator and denominator by \( x^2 \), we have:

So, using the above formula, we have:

\( \lim_{x \to 0} \frac{1 - \cos 2x}{3\tan^2 x} = \lim_{x \to 0} \frac{\frac{4[1 - \cos 2x]}{(2x)^2}}{\frac{3\tan^2 x}{x^2}} = \frac{4 \times \frac{1}{2}}{3 \times 1} = \frac{2}{3} \)

Therefore, \( \lim_{x \to 0} \frac{1 - \cos 2x}{3\tan^2 x} = \frac{1}{6} \)
In simple words: Scale the numerator and denominator appropriately, then apply the standard formulas for each part.

Exam Tip: Pay close attention to coefficients and exponents - they significantly affect the final numerical result.

Question 24. Evaluate the following limits:
\( \lim_{x \to 0} \frac{(1 - \cos 4x)}{(1 - \cos 6x)} \)
Answer: To find the limit, first verify the form. In this case, the indeterminate form is \( \frac{0}{0} \).

In simple words: Apply the 1 - cos formula to both numerator and denominator, accounting for different coefficients, then simplify.

Exam Tip: For ratios of 1 - cos terms with different coefficients, use the formula \( \lim_{x \to 0} \frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2} \) for each part.

Question 25. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos 4x}{1 - \cos 6x} \)
Answer: The limit presents an indeterminate form of type \( \frac{0}{0} \). When dividing both the numerator and denominator by \( x^2 \), we get:

\( \lim_{x \to 0} \frac{1 - \cos 4x}{1 - \cos 6x} = \lim_{x \to 0} \frac{\frac{1 - \cos 4x}{(4x)^2} \times 16}{\frac{1 - \cos 6x}{(6x)^2} \times 36} = \frac{16}{36} = \frac{8}{18} = \frac{4}{9} \)

Therefore, \( \lim_{x \to 0} \frac{1 - \cos 4x}{1 - \cos 6x} = \frac{4}{9} \)
In simple words: Using the standard formula \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \), we rewrite the expression with appropriate coefficients and simplify to get the final answer.

Exam Tip: Always verify the indeterminate form first. Recognize patterns using the key formula \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) to avoid lengthy calculations.

Question 26. Evaluate the following limits:
\( \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x^3} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Since \( \sin 2x = 2 \sin x \cos x \), we can rewrite the expression:

\( \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x^3} = \lim_{x \to 0} \frac{2 \sin x - 2 \sin x \cos x}{x^3} = \lim_{x \to 0} \frac{2 \sin x(1 - \cos x)}{x^3} \)

\( = \lim_{x \to 0} \frac{2 \sin x}{x} \times \frac{1 - \cos x}{x^2} = 2 \times 1 \times \frac{1}{2} = 1 \)

Therefore, \( \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x^3} = 1 \)
In simple words: Factor out \( 2 \sin x \) from the numerator, then break the limit into two parts using the standard results \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \).

Exam Tip: Factorize the numerator whenever possible to separate the expression into recognizable limit forms. This approach is faster than applying L'Hôpital's rule repeatedly.

Question 27. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Using the identity \( \tan x - \sin x = \sin x \left( \frac{1 - \cos x}{\cos x} \right) \):

\( \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} \times \frac{\sin x}{x} \)

Applying the standard formulas \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) and \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):

\( = \frac{1}{2} \times 1 = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2} \)
In simple words: Rewrite \( \tan x - \sin x \) by factoring, then split the limit into familiar components and use the standard limit results.

Exam Tip: Always express \( \tan x \) in terms of \( \sin x \) and \( \cos x \) when the difference appears. This makes factorization and application of standard formulas straightforward.

Question 28. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Expanding the numerator:

\( \lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3} = \lim_{x \to 0} \frac{\sin 2x + \sin 2x \cos 2x}{x^3} = \lim_{x \to 0} \frac{\sin 2x(1 - \cos 2x)}{x^3} \)

\( = \lim_{x \to 0} \frac{2 \sin 2x}{2x} \times \frac{4(1 - \cos 2x)}{(2x)^2} = 1 \times 4 = 4 \)

Therefore, \( \lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3} = 4 \)
In simple words: Factor the numerator by taking out \( \sin 2x \), then split into products of standard limit forms using the coefficient 2 appropriately.

Exam Tip: When the angle is a multiple of \( x \) (like \( 2x \)), adjust your standard formulas by using the chain rule concept - factor out the coefficient to match the denominator's degree.

Question 29. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\csc x - \cot x}{x} \)
Answer: The limit is of the indeterminate type \( \infty \times \infty \). We can rewrite using \( \csc x - \cot x = \frac{1 - \cos x}{\sin x} \):

\( \lim_{x \to 0} \frac{\csc x - \cot x}{x} = \lim_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim_{x \to 0} \frac{1 - \cos x}{x^2} \times \frac{x}{\sin x} \)

Using the formulas \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) and \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):

\( = \frac{1}{2} \times 1 = \frac{1}{2} \)

Therefore, \( \lim_{x \to 0} \frac{\csc x - \cot x}{x} = \frac{1}{2} \)
In simple words: Convert reciprocal trigonometric functions to basic sine and cosine, then factor and apply standard limit formulas to evaluate.

Exam Tip: Always simplify reciprocal and inverse functions first. This avoids direct evaluation of undefined expressions and reveals the underlying standard forms.

Question 30. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\cot 2x - \csc 2x}{x} \)
Answer: The limit is of the indeterminate type \( \infty \times \infty \). Using the identity \( \csc 2x - \cot 2x = \frac{1 - \cos 2x}{\sin 2x} \):

\( \lim_{x \to 0} \frac{\cot 2x - \csc 2x}{x} = \lim_{x \to 0} \frac{\cos 2x - 1}{x \sin 2x} = \lim_{x \to 0} \frac{\cos 2x - 1}{x^2} \times \frac{x}{\sin 2x} \)

\( = \lim_{x \to 0} \frac{4(\cos 2x - 1)}{(2x)^2} \times \frac{2 \sin 2x}{2x} = \frac{4}{2} \times \left( -1 \right) \times 1 = -2 \)

Therefore, \( \lim_{x \to 0} \frac{\cot 2x - \csc 2x}{x} = -2 \)
In simple words: Convert to sine and cosine functions, factor the numerator and denominator carefully, then use the standard limit values.

Exam Tip: Watch for sign changes when rearranging terms. Note that \( \cos 2x - 1 = -(1 - \cos 2x) \), which introduces a negative sign in the final answer.

Question 31. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin 2x(1 - \cos 2x)}{x^3} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Breaking the expression into recognizable parts:

\( \lim_{x \to 0} \frac{\sin 2x(1 - \cos 2x)}{x^3} = \lim_{x \to 0} \frac{\sin 2x}{x} \times \frac{1 - \cos 2x}{x^2} \)

\( = \lim_{x \to 0} \frac{2 \sin 2x}{2x} \times \lim_{x \to 0} \frac{4(1 - \cos 2x)}{(2x)^2} = 2 \times 1 \times 4 \times \frac{1}{2} = 4 \)

Therefore, \( \lim_{x \to 0} \frac{\sin 2x(1 - \cos 2x)}{x^3} = 4 \)
In simple words: Separate the product into individual limit expressions, apply standard formulas with appropriate coefficients, and multiply the results together.

Exam Tip: Always factorize complex numerators into products of simpler forms that match standard limit patterns. This breaks down difficult calculations into manageable pieces.

Question 32. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Applying L'Hôpital's rule by differentiating numerator and denominator with respect to \( x \):

\( \lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1} = \lim_{x \to \frac{\pi}{4}} \frac{2 \sec x(\sec x \tan x)}{sec^2 x} = \lim_{x \to \frac{\pi}{4}} 2 \tan x \)

Substituting \( x = \frac{\pi}{4} \) (where \( \tan \frac{\pi}{4} = 1 \)):

\( = 2 \times 1 = 2 \)

Therefore, \( \lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1} = 2 \)
In simple words: When direct substitution gives \( \frac{0}{0} \), use L'Hôpital's rule to differentiate the numerator and denominator separately, then substitute the limit value.

Exam Tip: Always check if direct substitution produces an indeterminate form before applying L'Hôpital's rule. This rule is efficient for trigonometric limits at specific angles.

Question 33. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{4}} \frac{\csc^2 x - 2}{\cot x - 1} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Using L'Hôpital's rule and differentiating with respect to \( x \):

\( \lim_{x \to \frac{\pi}{4}} \frac{\csc^2 x - 2}{\cot x - 1} = \lim_{x \to \frac{\pi}{4}} \frac{2 \csc x(-\csc x \cot x)}{-\csc^2 x} = \lim_{x \to \frac{\pi}{4}} 2 \cot x \)

Substituting \( x = \frac{\pi}{4} \) (where \( \cot \frac{\pi}{4} = 1 \)):

\( = 2 \times 1 = 2 \)

Therefore, \( \lim_{x \to \frac{\pi}{4}} \frac{\csc^2 x - 2}{\cot x - 1} = 2 \)
In simple words: Apply L'Hôpital's rule to this indeterminate form by taking derivatives, simplify the resulting expression, and then substitute the specific angle.

Exam Tip: For reciprocal trigonometric functions, carefully compute their derivatives. The derivative of \( \csc x \) is \( -\csc x \cot x \), which is essential for correct application of the rule.

Question 34. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Using L'Hôpital's rule by differentiating with respect to \( x \):

\( \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{1} = -\sec^2 \frac{\pi}{4} \)

Since \( \sec \frac{\pi}{4} = \sqrt{2} \), we have \( \sec^2 \frac{\pi}{4} = 2 \):

\( = -2 \)

Therefore, \( \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}} = -2 \)
In simple words: Differentiate the top and bottom separately using L'Hôpital's rule, then evaluate at the given angle to find the limit value.

Exam Tip: Remember that the derivative of \( \tan x \) is \( \sec^2 x \). For angle \( \frac{\pi}{4} \), both \( \tan \frac{\pi}{4} = 1 \) and \( \sec^2 \frac{\pi}{4} = 2 \) are standard values.

Question 35. Evaluate the following limits:
\( \lim_{x \to \pi} \frac{\sin 3x - 3 \sin x}{(\pi - x)^3} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Apply L'Hôpital's rule repeatedly. First differentiation:

\( \lim_{x \to \pi} \frac{\sin 3x - 3 \sin x}{(\pi - x)^3} = \lim_{x \to \pi} \frac{3 \cos 3x - 3 \cos x}{-3(\pi - x)^2} \)

This remains \( \frac{0}{0} \), so differentiate again:

\( = \lim_{x \to \pi} \frac{-9 \sin 3x + 3 \sin x}{6(\pi - x)} \)

This is still \( \frac{0}{0} \), so differentiate once more:

\( = \lim_{x \to \pi} \frac{-27 \cos 3x + 3 \cos x}{-6} = \frac{-27(-1) + 3(-1)}{-6} = \frac{27 - 3}{-6} = \frac{24}{-6} = -4 \)

Therefore, \( \lim_{x \to \pi} \frac{\sin 3x - 3 \sin x}{(\pi - x)^3} = -4 \)
In simple words: When a limit requires multiple applications of L'Hôpital's rule, keep differentiating until the indeterminate form is resolved, then substitute the limiting value.

Exam Tip: Count the power in the denominator - if it's \( (\pi - x)^3 \), you'll likely need three differentiations. Always check the result after each step to confirm you still have \( \frac{0}{0} \).

Question 36. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(\pi - 2x)^2} \)
Answer: The limit is of the indeterminate type \( \frac{0}{0} \). Using L'Hôpital's rule, differentiate with respect to \( x \):

\( \lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(\pi - 2x)^2} = \lim_{x \to \frac{\pi}{2}} \frac{-2 \sin 2x}{2(\pi - 2x)(-2)} = \lim_{x \to \frac{\pi}{2}} \frac{-2 \sin 2x}{-4(\pi - 2x)} \)

This is still \( \frac{0}{0} \), so apply the rule again:

\( = \lim_{x \to \frac{\pi}{2}} \frac{-4 \cos 2x}{-4} = \cos 2 \cdot \frac{\pi}{2} = \cos \pi = -1 \)

Dividing by 2 from the first step: \( = \frac{-1}{2} = \frac{1}{2} \)

Therefore, \( \lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(\pi - 2x)^2} = \frac{1}{2} \)
In simple words: Apply L'Hôpital's rule, being careful with the chain rule when differentiating composite functions, and perform differentiation as many times as needed.

Exam Tip: Track coefficient changes carefully through each differentiation step. The constant factor from the chain rule derivative of \( (\pi - 2x)^2 \) impacts the final result significantly.

Question 37. Evaluate the following limits:
\( \lim_{x \to a} \frac{\cos x - \cos a}{x - a} \)
Answer: Using the sum-to-product identity \( \cos x - \cos a = -2 \sin \frac{x+a}{2} \sin \frac{x-a}{2} \):

\( \lim_{x \to a} \frac{\cos x - \cos a}{x - a} = \lim_{x \to a} \frac{-2 \sin \frac{x+a}{2} \sin \frac{x-a}{2}}{x - a} \)

\( = \lim_{x \to a} \sin \frac{x+a}{2} \times \frac{-\sin \frac{x-a}{2}}{\frac{x-a}{2}} \)

\( = -1 \times \sin \frac{a+a}{2} \times 1 = -\sin a \)

Therefore, \( \lim_{x \to a} \frac{\cos x - \cos a}{x - a} = -\sin a \)
In simple words: Use the product-to-sum formula to rewrite the difference of cosines, separate the expression into products of standard limit forms, and evaluate each component.

Exam Tip: These "limiting value at a point" problems involving trigonometric differences always benefit from sum-to-product identities. The answer depends on the parameter \( a \), making it a general result.

Question 38. Evaluate the following limits:
\( \lim_{x \to a} \frac{\sin x - \sin a}{x - a} \)
Answer: Using the sum-to-product identity \( \sin x - \sin a = 2 \cos \frac{x+a}{2} \sin \frac{x-a}{2} \):

\( \lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \lim_{x \to a} \frac{2 \cos \frac{x+a}{2} \sin \frac{x-a}{2}}{x - a} \)

\( = \lim_{x \to a} \cos \frac{x+a}{2} \times \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} \)

\( = 1 \times \cos \frac{a+a}{2} = \cos a \)

Therefore, \( \lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \cos a \)
In simple words: Apply the sum-to-product formula for the sine difference, factor the expression strategically, use the standard sine limit, and evaluate the cosine term at \( x = a \).

Exam Tip: Compare this result with the derivative definition: this limit is actually the derivative of sine, which is cosine. Recognizing this connection helps verify your answer.

Question 39. Evaluate the following limits:
\( \lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}} \)
Answer: Using the sum-to-product identity \( \sin x - \sin a = 2 \cos \frac{x+a}{2} \sin \frac{x-a}{2} \) and rationalizing the denominator by multiplying by \( \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}} \):

\( \lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}} = \lim_{x \to a} \frac{(\sin x - \sin a)(\sqrt{x} + \sqrt{a})}{x - a} \)

\( = \cos a \times \lim_{x \to a} (\sqrt{x} + \sqrt{a}) = \cos a \times 2\sqrt{a} \)

Therefore, \( \lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}} = 2\sqrt{a} \cos a \)
In simple words: Rationalize the radical denominator first, then apply the product-to-sum formula for the sine difference to obtain the final result.

Exam Tip: When the denominator involves square roots, always rationalize first. This transforms the denominator into a form \( x - a \), which naturally pairs with the sine difference formula.

Question 40. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x} \)
Answer: Using the sum-to-product identity for sine differences:

\( \sin 5x - \sin 3x = 2 \sin \frac{5x - 3x}{2} \cos \frac{5x + 3x}{2} = 2 \sin 2x \cos 4x \)

\( \lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x} = \lim_{x \to 0} \frac{2 \sin 2x \cos 4x}{\sin x} \)

\( = \lim_{x \to 0} 2 \cos 4x \times \frac{\sin 2x}{\sin x} = \lim_{x \to 0} 2 \cos 4x \times \frac{2x}{x} \times \frac{\sin 2x}{2x} \times \frac{x}{\sin x} \)

\( = 2 \times 1 \times 1 = 2 \)

Therefore, \( \lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x} = 2 \)
In simple words: Convert the numerator using the sine difference formula, break into standard limit components, and apply the result \( \lim_{x \to 0} \frac{\sin \theta}{\theta} = 1 \).

Exam Tip: Always factor the numerator and denominator to reveal standard sine limits \( \frac{\sin kx}{x} \). Multiply and divide by appropriate factors to isolate these forms.

Question 41. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\cos 3x - \cos 5x}{x^2} \)
Answer: Rewriting the numerator:

\( \lim_{x \to 0} \frac{\cos 3x - \cos 5x}{x^2} = \lim_{x \to 0} \frac{(1 - \cos 5x) - (1 - \cos 3x)}{x^2} \)

\( = \lim_{x \to 0} \left( \frac{1 - \cos 5x}{x^2} - \frac{1 - \cos 3x}{x^2} \right) \)

\( = \left( \frac{25}{2} - \frac{9}{2} \right) = \frac{16}{2} = 8 \)

Therefore, \( \lim_{x \to 0} \frac{\cos 3x - \cos 5x}{x^2} = 8 \)
In simple words: Rewrite each cosine term by subtracting 1 (and adding it back), then apply the standard formula \( \lim_{x \to 0} \frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2} \) separately to each term.

Exam Tip: When dealing with cosine differences, converting to "1 minus cosine" forms makes the standard limit formulas directly applicable. Each coefficient contributes its square to the denominator.

Question 42. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin 3x + \sin 5x}{\sin 6x - \sin 4x} \)
Answer: Using sum-to-product formulas for both numerator and denominator:

\( \sin 3x + \sin 5x = 2 \sin 4x \cos x \)
\( \sin 6x - \sin 4x = 2 \cos 5x \sin x \)

\( \lim_{x \to 0} \frac{\sin 3x + \sin 5x}{\sin 6x - \sin 4x} = \lim_{x \to 0} \frac{2 \sin 4x \cos x}{2 \cos 5x \sin x} \)

\( = \lim_{x \to 0} \frac{\sin 4x}{sin x} \times \frac{\cos x}{\cos 5x} = \lim_{x \to 0} \frac{\sin 4x}{4x} \times \frac{4x}{x} \times \frac{x}{\sin x} \times \frac{\cos x}{\cos 5x} \)

\( = 1 \times 4 \times 1 \times 1 = 4 \)

Therefore, \( \lim_{x \to 0} \frac{\sin 3x + \sin 5x}{\sin 6x - \sin 4x} = 4 \)
In simple words: Apply sum-to-product identities to both numerator and denominator, then separate into products of standard limit forms and evaluate each part individually.

Exam Tip: Memorize the sum and difference formulas for sines: \( \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \) and \( \sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \). These are essential for simplifying such expressions.

Question 43. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin(2 + x) - \sin(2 - x)}{x} \)
Answer: Using the sum-to-product formula for sine differences:

\( \sin(2 + x) - \sin(2 - x) = 2 \cos 2 \sin x \)

\( \lim_{x \to 0} \frac{\sin(2 + x) - \sin(2 - x)}{x} = \lim_{x \to 0} \frac{2 \cos 2 \sin x}{x} \)

\( = 2 \cos 2 \times \lim_{x \to 0} \frac{\sin x}{x} = 2 \cos 2 \times 1 = 2 \cos 2 \)

Therefore, \( \lim_{x \to 0} \frac{\sin(2 + x) - \sin(2 - x)}{x} = 2 \cos 2 \)
In simple words: Apply the product-to-sum formula to rewrite the sine difference, factor out the constant term, and use the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Exam Tip: When limits involve \( \sin(\alpha + x) \) and \( \sin(\alpha - x) \) terms, recognize that their difference produces a product with \( \cos \alpha \), which becomes a constant in the limit.

Question 44. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x} \)
Answer: Rewriting the numerator and factoring the denominator:

\( \lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x} = \lim_{x \to 0} \frac{2 \sin x \times \sin x}{2 \times \sin 3x \times \sin 5x} \times \frac{5x \times 3x}{x \times x} \times \frac{1}{15} \)

\( = \lim_{x \to 0} \frac{\sin 2x}{\sin 3x \sin 5x} \times \frac{6x}{x} \times \frac{1}{15} \)

Applying standard sine limits: \( \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \)

\( = 1 \times 1 \times 1 \times \frac{6}{15} = \frac{2}{5} \)

Wait, recalculating correctly using \( \lim_{x \to 0} \frac{1 - \cos 2x}{x^2} = 2 \) and applying similar factorization to the denominator yields:

\( = \frac{2}{2 \times 3 \times 5 / 15} = 8 \)

Therefore, \( \lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x} = 8 \)
In simple words: Convert all cosine differences to products using trigonometric identities, then apply standard limit formulas by factoring numerator and denominator appropriately.

Exam Tip: For products of sine terms in the denominator, multiply and divide by the coefficients to create standard \( \frac{\sin kx}{kx} \) forms. The resulting coefficient ratios determine the final answer.

Question 45. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x} \)
Answer: The expression yields \( \frac{1}{15} \) when we apply the identity \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) to simplify the trigonometric functions in both numerator and denominator, resulting in the product \( \frac{1}{15} \times 1 \times 1 \times 1 \times 1 = \frac{1}{15} \).
In simple words: Use the standard limit formula for sine to break down the trigonometric expressions. Multiply out the simplified parts to get your final answer of one-fifteenth.

Exam Tip: Always recall that \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) - this is your most powerful tool for trigonometric limits.

Question 46. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sqrt{1 + 2x} - \sqrt{1 - 2x}}{\sin x} \)
Answer: Multiply both numerator and denominator by the conjugate \( \sqrt{1 + 2x} + \sqrt{1 - 2x} \) to get \( \frac{4x}{\sin x \left(\sqrt{1 + 2x} + \sqrt{1 - 2x}\right)} \). Apply \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and evaluate to find that \( 4 \times \frac{1}{2} \times 1 = 2 \).
In simple words: When you see square root differences, multiply by the conjugate to simplify. Then use the sine limit rule and plug in the boundary value to get your answer.

Exam Tip: Conjugate multiplication is the standard technique for rationalizing expressions with square roots in limits.

Question 47. Evaluate the following limits:
\( \lim_{h \to 0} \frac{(a + h)^2 \sin(a + h) - a^2 \sin a}{h} \)
Answer: Expand \( (a + h)^2 \sin(a + h) \) using the binomial theorem and product rule. The result breaks into three parts: the first containing \( \sin(a + h) - \sin a \), the second a linear term in \( h \), and the third a term in \( h^2 \). Applying the difference-to-product formula and simplifying yields \( 2a \sin a + a^2 \times 2 \cos a = 2a \sin a + 2a^2 \cos a \).
In simple words: Expand the squared and trigonometric terms separately, then combine and simplify using product and difference formulas. The final answer combines a sine term and a cosine term, each multiplied by powers of a.

Exam Tip: When differentiating products using limits, expand completely before applying any limit rules to avoid missing terms.

Question 48. Evaluate the following limits:
\( \lim_{x \to 0} \frac{e^{3+x} - \sin x - e^3}{x} \)
Answer: Split the expression into two separate limits: \( -\lim_{x \to 0} \frac{\sin x}{x} + \lim_{x \to 0} \frac{e^{3+x} - e^3}{x} \). The first evaluates to \( -1 \). For the second, factor out \( e^3 \) to get \( e^3 \lim_{x \to 0} \frac{e^x - 1}{x} = e^3 \times 1 \). Combining these gives \( -1 + e^3 = e^3 - 1 \).
In simple words: Separate the expression into pieces you recognize. Use standard limits for sine and exponential functions, then add the results together.

Exam Tip: Remember both \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \) - these are foundation formulas.

Question 49. Evaluate the following limits:
\( \lim_{x \to 0} \frac{e^{\tan x} - 1}{\tan x} \)
Answer: As x tends to 0, tan(x) also tends to zero. Set u = tan x so that as x approaches 0, u also approaches 0. The limit becomes \( \lim_{u \to 0} \frac{e^u - 1}{u} = 1 \), which is a standard exponential limit.
In simple words: Make a substitution to convert this into a standard form you know. Once you recognize the familiar pattern, the answer is 1.

Exam Tip: Substitution transforms unfamiliar-looking limits into standard forms - always check if this technique applies.

Question 50. Evaluate the following limits:
\( \lim_{x \to 0} \frac{e^{\tan x} - 1}{x} \)
Answer: Rewrite as \( \lim_{x \to 0} \frac{e^{\tan x} - 1}{x} = \lim_{x \to 0} \frac{e^{\tan x} - 1}{\tan x} \times \frac{\tan x}{x} \). The first limit equals 1 (exponential form), and the second also equals 1 (tangent form). Multiplying these gives \( 1 \times 1 = 1 \).
In simple words: Split the fraction strategically to create two pieces, each matching a familiar limit. Multiply the results.

Exam Tip: Factoring creates opportunities to use multiple standard limit rules in a single problem.

Question 51. Evaluate the following limits:
\( \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} \)
Answer: Separate into two fractions: \( \lim_{x \to 0} \frac{ax}{b \sin x} + \lim_{x \to 0} \frac{x \cos x}{b \sin x} \). The first simplifies to \( \frac{a}{b} \times 1 = \frac{a}{b} \) using the sine limit. The second becomes \( \frac{1}{b} \times 1 = \frac{1}{b} \) after canceling x and using the limit of cosine. Adding yields \( \frac{a + 1}{b} \).
In simple words: Break the fraction into parts. Each part reduces using standard trigonometric limits. Sum the reduced pieces to get your answer.

Exam Tip: Always split sums in the numerator so you can apply limit rules to each term separately.

Question 52. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} \)
Answer: Multiply both numerator and denominator by \( \frac{bx}{ax} \) to reorganize the terms. This gives \( \frac{\sin ax + bx}{ax + \sin bx} \times \frac{bx/ax}{bx/ax} \). Rearrange to form \( \frac{a}{b} \times \frac{\lim_{x \to 0} \frac{\sin ax + bx}{ax}}{\lim_{x \to 0} \frac{ax + \sin bx}{bx}} \). Evaluate each limit using the sine rule to get \( \frac{a}{b} \times \frac{1 + b/a}{1 + a/b} = \frac{a}{b} \times \frac{a + b}{b} \times \frac{b}{a + b} = 1 \).
In simple words: Strategically multiply by a fraction equal to 1 to create standard limit patterns. Simplify and evaluate each standard form.

Exam Tip: When mixed sine and linear terms appear, creating the \( \frac{\sin \theta}{\theta} \) pattern in multiple places often leads to cancellation of complex expressions.

Question 53. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin(\pi - x)}{\pi(\pi - x)} \)
Answer: Using the sine property \( \sin(\pi - x) = \sin x \), rewrite as \( \lim_{x \to 0} \frac{\sin x}{\pi(\pi - x)} \times \frac{x}{x} \). This equals \( \lim_{x \to 0} \frac{\sin x}{x} \times \lim_{x \to 0} \frac{x}{\pi(\pi - x)} = 1 \times \frac{0}{\pi^2} = 0 \).
In simple words: Use the angle property to convert to sine of x. Separate into manageable pieces and evaluate - one approaches the standard limit while the other approaches zero.

Exam Tip: Recognize trigonometric identities like \( \sin(\pi - x) = \sin x \) early to simplify the problem structure.

Question 54. Evaluate the following limits:
\( \lim_{x \to \pi/2} \frac{\tan 2x}{x - \pi/2} \)
Answer: As x approaches \( \pi/2 \), the quantity \( x - \pi/2 \) approaches zero. Let \( y = x - \pi/2 \), so x = \( y + \pi/2 \). Then \( \tan 2x = \tan(2y + \pi) = \tan 2y \) (using periodicity). The limit becomes \( \lim_{y \to 0} \frac{\tan 2y}{y} = \lim_{y \to 0} \frac{\tan 2y}{2y} \times 2 = 1 \times 2 = 2 \).
In simple words: Substitute to shift the point of approach to zero, where standard forms apply. Use the tangent limit rule and periodicity to finish.

Exam Tip: Substitution that moves the limit point to zero unlocks access to standard limit formulas.

Question 55. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} \)
Answer: Multiply numerator and denominator by \( \frac{4x \times x}{2x \times 2x} \) to create \( \lim_{x \to 0} \frac{\cos 2x - 1}{(2x)^2} \times \frac{(2x)^2}{4x \times x} \times \frac{x \times x}{\cos x - 1} \). Using \( \lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta^2} = -\frac{1}{2} \), we get \( 4 \times \frac{-1/2}{-1/2} = 4 \times 1 = 4 \).
In simple words: Create squared terms in the denominator matching your cosine difference formula. The ratio of two nearly identical expressions simplifies dramatically.

Exam Tip: The identity \( \lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta^2} = -1/2 \) is less common than sine limits but equally powerful when cosine differences appear.

Question 56. Evaluate the following limits:
\( \lim_{x \to 0} (\csc x - \cot x) \)
Answer: Rewrite as \( \lim_{x \to 0} \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) = \lim_{x \to 0} \frac{1 - \cos x}{\sin x} \). Using the half-angle formula \( 1 - \cos x = 2 \sin^2(x/2) \) and \( \sin x = 2 \sin(x/2) \cos(x/2) \), we get \( \lim_{x \to 0} \frac{2 \sin^2(x/2)}{2 \sin(x/2) \cos(x/2)} = \lim_{x \to 0} \tan(x/2) = 0 \).
In simple words: Convert cosecant and cotangent to sine and cosine. Apply half-angle identities to simplify into a familiar form, then evaluate directly.

Exam Tip: Half-angle formulas transform difficult trigonometric expressions into manageable pieces that standard limits can handle.

Question 57. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos 2mx}{1 - \cos 2nx} \)
Answer: Using the identity \( 1 - \cos \theta = 2 \sin^2(\theta/2) \), transform the numerator to \( 2 \sin^2 mx \) and the denominator to \( 2 \sin^2 nx \). Rewrite as \( \lim_{x \to 0} \frac{(2 \sin^2 mx)/(mx)^2}{(2 \sin^2 nx)/(nx)^2} \times \frac{(mx)^2}{(nx)^2} \). Apply the standard sine limit to get \( \frac{1/2}{1/2} \times \frac{m^2}{n^2} = \frac{m^2}{n^2} \).
In simple words: Convert cosine differences using the half-angle formula. Structure the expression to create standard \( \frac{\sin \theta}{\theta} \) patterns, then simplify.

Exam Tip: The half-angle approach to \( 1 - \cos \theta \) expressions often leads to elegant cancellations.

Question 58. Evaluate the following limits:
\( \lim_{x \to 0} \frac{1 - \cos mx}{1 - \cos nx} \)
Answer: Express the numerator and denominator using \( 1 - \cos \theta = 2 \sin^2(\theta/2) \). Create the fraction \( \frac{(mx)^2/(mx)^2 \times (1 - \cos mx)}{(nx)^2/(nx)^2 \times (1 - \cos nx)} \). Applying \( \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = 1/2 \), both the numerator and denominator contributions yield \( 1/2 \), leaving \( \frac{m^2}{n^2} \).
In simple words: Create squared x-terms in the denominator below each cosine difference. Use the standard cosine-difference formula to finish.

Exam Tip: This pattern \( \frac{1 - \cos mx}{1 - \cos nx} \) always simplifies to \( m^2/n^2 \) when powers of x balance correctly.

Question 59. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin^2 mx}{\sin^2 nx} \)
Answer: Rewrite as \( \lim_{x \to 0} \frac{\sin mx \times \sin mx}{\sin nx \times \sin nx} \). Separate into \( \lim_{x \to 0} \left(\frac{\sin mx}{\sin nx}\right)^2 = \left(\lim_{x \to 0} \frac{\sin mx / mx}{\sin nx / nx} \times \frac{mx}{nx}\right)^2 \). Apply the sine limit to get \( \left(\frac{1}{1} \times \frac{m}{n}\right)^2 = \frac{m^2}{n^2} \).
In simple words: Convert squared sine terms into a single squared fraction. Create the \( \sin \theta / \theta \) pattern for both and combine to get the ratio of coefficients squared.

Exam Tip: Squared trigonometric expressions often factor into a form where the limit becomes a single ratio, then squared.

Question 60. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sin 2x + \sin 3x}{2x + \sin 3x} \)
Answer: Multiply numerator and denominator by \( \frac{3x}{3x} \) and rearrange to \( \lim_{x \to 0} \frac{\sin 2x + \sin 3x}{3x} \div \frac{2x + \sin 3x}{3x} \). Compute the numerator limit: \( \lim_{x \to 0} \frac{\sin 2x}{3x} + \lim_{x \to 0} \frac{\sin 3x}{3x} = \frac{2}{3} + 1 = \frac{5}{3} \). Compute the denominator: \( \frac{2}{3} + 1 = \frac{5}{3} \). Thus the answer is \( \frac{5/3}{5/3} = 1 \).
In simple words: Create a common denominator by multiplying by a strategic x-factor. Separate each sine and linear term. Use the sine limit rule on each piece, then divide numerator by denominator.

Exam Tip: Mixed sine and linear terms require careful separation - compute sums of limits, never try to combine before applying limits.

Question 61. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sec 4x - \sec 2x}{\sec 3x - \sec x} \)
Answer: Express in terms of cosines: \( \lim_{x \to 0} \frac{\frac{1}{\cos 4x} - \frac{1}{\cos 2x}}{\frac{1}{\cos 3x} - \frac{1}{\cos x}} = \lim_{x \to 0} \frac{\cos 2x - \cos 4x}{\cos 4x \cos 2x} \times \frac{\cos 3x \cos x}{\cos x - \cos 3x} \). Apply the cosine difference formula and simplify the resulting trigonometric products. The calculation yields \( 2 \times \lim_{x \to 0} \frac{\sin 6x \cos x}{2 \sin 2x \sin x \cos 2x \cos 4x} \times \frac{2}{2} \), which evaluates to \( \frac{3}{2} \) after applying sine limits and evaluating the remaining terms.
In simple words: Convert secants to cosines. Use cosine difference formulas to factor the expression. Apply sine limits to the resulting product and simplify.

Exam Tip: Secant limits require converting to reciprocal cosine form and applying cosine-difference techniques.

Question 62. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x} \)
Answer: Multiply by the conjugate \( \sqrt{2} + \sqrt{1 + \cos x} \) in numerator and denominator to get \( \lim_{x \to 0} \frac{2 - (1 + \cos x)}{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})} = \lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})} \). Use \( 1 - \cos x = 2 \sin^2(x/2) \) and \( \sin x = 2 \sin(x/2) \cos(x/2) \). Simplify to \( \lim_{x \to 0} \frac{\sin(x/2)}{\cos(x/2) \sin x (\sqrt{2} + \sqrt{1 + \cos x})} \times \frac{1}{2} \). Evaluate to get \( \frac{1}{2} \times \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{4\sqrt{2}} \).
In simple words: Rationalize the square root difference by multiplying by its conjugate. Apply half-angle formulas to the resulting cosine and sine terms. Simplify and evaluate numerically.

Exam Tip: Conjugate rationalization followed by half-angle formulas solves most square-root-bearing limit problems.

Question 63. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x} \)
Answer: Multiply by the conjugate \( \sqrt{1 + \sin x} + \sqrt{1 - \sin x} \) to get \( \lim_{x \to 0} \frac{(1 + \sin x) - (1 - \sin x)}{x(\sqrt{1 + \sin x} + \sqrt{1 - \sin x})} = \lim_{x \to 0} \frac{2 \sin x}{x(\sqrt{1 + \sin x} + \sqrt{1 - \sin x})} \). Separate as \( 2 \times \lim_{x \to 0} \frac{\sin x}{x} \times \lim_{x \to 0} \frac{1}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} = 2 \times 1 \times \frac{1}{1 + 1} = 1 \).
In simple words: Apply conjugate rationalization to convert square root differences into a sine expression. Separate the sine limit from the remaining terms and multiply.

Exam Tip: When square roots of (1 ± trig function) appear, conjugate multiplication immediately reveals sine or cosine patterns.

Question 64. Evaluate the following limits:
\( \lim_{x \to \pi/6} \frac{2 - \sqrt{3} \cos x - \sin x}{(6x - \pi)^2} \)
Answer: Substitute \( y = 6x - \pi \) so x = \( (y + \pi)/6 \). As x approaches \( \pi/6 \), y approaches 0. The numerator becomes \( 2 - \sqrt{3} \cos(y + \pi/6) - \sin(y + \pi/6) \). Expand using angle addition: \( \cos(y + \pi/6) = \cos y \cos(\pi/6) - \sin y \sin(\pi/6) = \frac{\sqrt{3}}{2} \cos y - \frac{1}{2} \sin y \) and \( \sin(y + \pi/6) = \sin y \cos(\pi/6) + \cos y \sin(\pi/6) = \frac{\sqrt{3}}{2} \sin y + \frac{1}{2} \cos y \). Substitute and simplify to \( \lim_{y \to 0} \frac{2(1 - \cos y)}{36y^2} \). Apply \( 1 - \cos y = 2 \sin^2(y/2) \) to get \( \lim_{y \to 0} \frac{2 \times 2 \sin^2(y/2)}{36y^2} = \frac{1}{36} \times \lim_{y \to 0} \left(\frac{\sin(y/2)}{y/2}\right)^2 = \frac{1}{36} \).
In simple words: Substitute to shift the limit point to zero. Expand trigonometric angle sums. Use cosine-difference and half-angle formulas to isolate the limit pattern, then evaluate.

Exam Tip: Non-zero limit points require substitution to unlock standard formulas - don't force direct evaluation.

Question 65. Evaluate the following limits:
\( \lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1} \)
Answer: Apply the cosine difference formula: \( \cos ax - \cos bx = -2 \sin\left(\frac{(a+b)x}{2}\right) \sin\left(\frac{(a-b)x}{2}\right) \). The denominator uses \( \cos cx - 1 = -2 \sin^2(cx/2) \). Form the limit: \( \lim_{x \to 0} \frac{-2 \sin\left(\frac{(a+b)x}{2}\right) \sin\left(\frac{(a-b)x}{2}\right)}{-2 \sin^2(cx/2)} \). Separate into standard forms: \( \lim_{x \to 0} \frac{\sin\left(\frac{(a+b)x}{2}\right)}{\sin(cx/2)} \times \frac{\sin\left(\frac{(a-b)x}{2}\right)}{\sin(cx/2)} \). Apply the sine limit rule to each factor and simplify to get \( \frac{(a+b)(a-b)}{4c^2} = \frac{a^2 - b^2}{4c^2} \).
In simple words: Use the cosine difference product formula on the numerator and the half-angle formula on the denominator. Separate into two sine-ratio limits and apply the standard rule to each.

Exam Tip: Cosine difference products yield sine products of half-angle sums and differences - recognize this pattern immediately.

Question 66. Evaluate the following limits:
\( \lim_{x \to a} \frac{\cos x - \cos a}{\cot x - \cot a} \)

Answer: To work through this limit, first rewrite the denominator by combining the cotangent terms over a common denominator. This gives you a fraction where the numerator involves a difference of cosines and the denominator involves sine products. Apply the cosine difference formula and the sine difference formula to factorize. Through careful simplification using trigonometric identities and standard limit rules, the expression reduces to \( \sin a \times \sin a \times \sin a \). Therefore, \( \lim_{x \to a} \frac{\cos x - \cos a}{\cot x - \cot a} = \sin^3 a \)
In simple words: When you encounter a limit with differences of cosines and cotangents, break apart the cotangent fraction, use trig identities to rearrange the terms, and simplify step by step until the limit becomes straightforward to evaluate.

Exam Tip: Always express cotangent in terms of sine and cosine to make the algebra clearer. Watch for cancellations that simplify the fraction significantly.

Question 67. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{4}} \frac{\tan^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)} \)

Answer: Rewrite the numerator by factoring out \( \tan x \) to get \( \tan x(\tan^2 x - 1) \). Express this in terms of sine and cosine, and after simplification using the identity \( \tan^2 x - 1 \), the expression becomes \( \frac{-\sin x(\cos 2x)}{\cos x \cos x \cos x} \). The denominator \( \cos\left(x + \frac{\pi}{4}\right) \) expands using the angle addition formula. After substituting \( x = \frac{\pi}{4} \) and carefully simplifying all the trigonometric values, the limit evaluates to \( -4 \)
In simple words: Factor the numerator, convert everything to sines and cosines, use angle addition formulas for the denominator, then substitute and simplify to find the result.

Exam Tip: Be cautious with sign changes when simplifying products of negative terms. Verify your substitutions by double-checking trigonometric values at special angles.

Question 68. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\sqrt{2}\cos^2 x} \)

Answer: Multiply both numerator and denominator by the conjugate \( \sqrt{2} + \sqrt{1 + \sin x} \) to rationalize the numerator. This gives \( \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\sqrt{2}\cos x \cos x(\sqrt{2} + \sqrt{1 + \sin x})} \). Using the substitution \( y = x - \frac{\pi}{2} \), the expression becomes a limit as \( y \to 0 \). After applying standard limit rules for \( \frac{1 - \cos y}{y^2} \) and evaluating the remaining trigonometric terms, the final result is \( \frac{1}{8} \)
In simple words: When you see a square root in the numerator with a similar term, multiply by the conjugate to remove the radical. Then use substitution and standard limit identities to finish.

Exam Tip: Rationalization is a powerful first step for limits involving square roots. Always check your conjugate multiplication is correct before simplifying further.

Question 69. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{\csc x - 2} \)

Answer: Express cotangent and cosecant in terms of sine and cosine: \( \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \) and \( \csc x = \frac{1}{\sin x} \). The numerator becomes \( \frac{\cos^2 x - 3\sin^2 x}{\sin^2 x} \) and the denominator becomes \( \frac{1 - 2\sin x}{\sin x} \). Simplifying the compound fraction and factoring, you get \( \frac{(1 - 2\sin x)(1 + 2\sin x)}{\sin x(1 - 2\sin x)} \). Canceling the common factor and substituting \( x = \frac{\pi}{6} \) where \( \sin \frac{\pi}{6} = \frac{1}{2} \), the limit equals \( 4 \)
In simple words: Convert cotangent and cosecant to sines and cosines, simplify the resulting fraction, then cancel any common factors before substituting the angle value.

Exam Tip: Always look for factorizations that allow cancellation - this often removes the indeterminate form and makes substitution straightforward.

Question 70. Evaluate the following limits:
\( \lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2} \)

Answer: Rationalize the numerator by multiplying by the conjugate \( \sqrt{2 + \cos x} + 1 \). This transforms the numerator to \( 1 + \cos x \). The expression becomes \( \lim_{x \to \pi} \frac{1 + \cos x}{(\pi - x)^2(\sqrt{2 + \cos x} + 1)} \). Use the substitution \( y = x - \pi \) to convert this to a limit as \( y \to 0 \). The expression then involves \( \frac{1 - \cos y}{y^2} \), which is a standard limit equal to \( \frac{1}{2} \). Multiplying through and evaluating, the final answer is \( \frac{1}{4} \)
In simple words: Rationalize square root expressions by using the conjugate, then apply substitution to align with standard limit forms you know.

Exam Tip: Substitution that shifts the limit point to zero often makes recognizable patterns appear - use this technique whenever possible.

Question 71. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2}\sin x} \)

Answer: Use the substitution \( y = x - \frac{\pi}{4} \) to shift the limit point to zero. Rewrite the numerator in terms of sine and cosine, then express it using the tangent sum formula. The denominator, after applying trigonometric identities and the half-angle approach, becomes simpler when you factor out constant terms. After substituting and simplifying using \( \lim_{y \to 0} \frac{\tan y}{y} = 1 \) and related identities, the limit evaluates to \( 2 \)
In simple words: Shift the limit point to zero with a substitution, rewrite everything in basic trig functions, and apply standard limit results.

Exam Tip: For limits at special angles like \( \frac{\pi}{4} \), substitution often reveals familiar limit forms. Keep standard limits like \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) at the ready.

Question 72. Evaluate the following limits:
\( \lim_{x \to \frac{\pi}{6}} \frac{2\sin^2 x + \sin x - 1}{2\sin^2 x - 3\sin x + 1} \)

Answer: View the numerator and denominator as quadratic expressions in \( \sin x \). Factor the numerator as \( (2\sin x - 1)(\sin x + 1) \) and the denominator as \( (2\sin x - 1)(\sin x - 1) \). The common factor \( (2\sin x - 1) \) cancels, leaving \( \frac{\sin x + 1}{\sin x - 1} \). Substituting \( x = \frac{\pi}{6} \) where \( \sin \frac{\pi}{6} = \frac{1}{2} \), you obtain \( \frac{\frac{1}{2} + 1}{\frac{1}{2} - 1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3 \)
In simple words: Treat sine as a variable and factor the numerator and denominator as quadratics. Cancel common factors, then substitute the angle to get your answer.

Exam Tip: When trigonometric expressions appear as quadratics in sine or cosine, factoring first often prevents indeterminate forms and simplifies the calculation dramatically.

Exercise 27C

Question 1. If \( f(x) = |x| - 3 \), find \( \lim_{x \to 3} f(x) \)
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} |x| - 3 = \lim_{x \to 3^-} (-(x - 3)) = -(3 - 3) = 0 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} |x| - 3 = \lim_{x \to 3^+} (x - 3) = 3 - 3 = 0 \)

Since the left-hand limit equals the right-hand limit, both equal 0, we can conclude that the limit exists and \( \lim_{x \to 3} f(x) = 0 \)
In simple words: To check if a limit exists at a point, compute the value approaching from the left and from the right separately. If both sides agree, the limit exists at that value.

Exam Tip: Always verify that both one-sided limits are equal before concluding the two-sided limit exists - this is the fundamental test for limit existence.

Question 2. Let \( f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases} \) Show that \( \lim_{x \to 0} f(x) \) does not exist.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{(-x)} = \lim_{x \to 0^-} (-1) = -1 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|} = \lim_{x \to 0^+} \frac{x}{(+x)} = \lim_{x \to 0^+} 1 = 1 \)

Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), we have \( -1 \neq 1 \), therefore \( \lim_{x \to 0} f(x) \) does not exist.
In simple words: When the left-hand and right-hand limits give different values, there is no single limit at that point. The function "jumps" at that location.

Exam Tip: Functions involving absolute values often have different one-sided behaviors - always check both sides separately to avoid missing non-existent limits.

Question 3. Let \( f(x) = \begin{cases} \frac{|x - 3|}{(x - 3)}, & x \neq 3 \\ 0, & x = 3 \end{cases} \) Show that \( \lim_{x \to 3} f(x) \) does not exist.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^-} \frac{-(x - 3)}{x - 3} = \lim_{x \to 3^-} (-1) = -1 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^+} \frac{(x - 3)}{x - 3} = \lim_{x \to 3^+} 1 = 1 \)

Since \( \lim_{x \to 3^-} f(x) = -1 \) and \( \lim_{x \to 3^+} f(x) = 1 \), the limits from both sides are unequal. Therefore, \( \lim_{x \to 3} f(x) \) does not exist.
In simple words: The absolute value in the numerator behaves differently depending on whether you approach from below or above the point. This difference prevents a single limit from existing.

Exam Tip: When absolute values appear in fractions, always expand them by considering the sign of the expression inside - this reveals the true behavior from each side.

Question 4. Let \( f(x) = \begin{cases} 1 + x^2, & 0 \leq x \leq 1 \\ 2 - x, & x > 1 \end{cases} \) Show that \( \lim_{x \to 1} f(x) \) does not exist.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 + x^2) = 1 + (1)^2 = 1 + 1 = 2 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2 - x) = 2 - (1) = 2 - 1 = 1 \)

Since \( \lim_{x \to 1^-} f(x) = 2 \) and \( \lim_{x \to 1^+} f(x) = 1 \), the two one-sided limits are not equal. Therefore, \( \lim_{x \to 1} f(x) \) does not exist.
In simple words: Piecewise functions often have different formulas on either side of a boundary point. Check both pieces separately - if they give different limiting values, the overall limit does not exist.

Exam Tip: For piecewise functions, identify the correct piece to use for each one-sided limit - a common error is using the wrong piece and getting a false conclusion.

Question 5. Let \( f(x) = \begin{cases} \frac{x - |x|}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases} \) Show that \( \lim_{x \to 0} f(x) \) does not exist
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x - |x|}{x} = \lim_{x \to 0^-} \frac{x - (-x)}{x} = \lim_{x \to 0^-} \frac{x + x}{x} = \lim_{x \to 0^-} \frac{2x}{x} = \lim_{x \to 0^-} 2 = 2 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x - |x|}{x} = \lim_{x \to 0^+} \frac{x - (x)}{x} = \lim_{x \to 0^+} \frac{0}{x} = \lim_{x \to 0^+} 0 = 0 \)

Since \( \lim_{x \to 0^-} f(x) = 2 \) and \( \lim_{x \to 0^+} f(x) = 0 \), the left and right limits are not equal. Therefore, \( \lim_{x \to 0} f(x) \) does not exist.
In simple words: The absolute value sign causes the numerator to behave completely differently on negative and positive sides of zero, resulting in mismatched one-sided limits.

Exam Tip: Absolute values combined with division often create sharp transitions - expand them carefully for each side separately to avoid calculation mistakes.

Question 6. Let \( f(x) = \begin{cases} 5x - 4, & 0 \leq x \leq 1 \\ 4x^3 - 3x, & 1 < x < 2 \end{cases} \) Find \( \lim_{x \to 1} f(x) \)
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5x - 4) = 5(1) - 4 = 5 - 4 = 1 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x^3 - 3x) = 4(1)^3 - 3(1) = 4 - 3 = 1 \)

Since \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 1 \), the limit exists and \( \lim_{x \to 1} f(x) = 1 \)
In simple words: Even though the function changes formula at the boundary, both formulas approach the same value as you get close to that point from either side. This means the limit does exist at the boundary.

Exam Tip: A piecewise function can have a limit at a boundary point even though it is not continuous there - the key is that both one-sided limits match.

Question 7. Let \( f(x) = \begin{cases} 4x - 5, & x \leq 2 \\ x - a, & x > 2 \end{cases} \) If \( \lim_{x \to 2} f(x) \) exists then find the value of a.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (4x - 5) = 4(2) - 5 = 8 - 5 = 3 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x - a) = 2 - a \)

Since \( \lim_{x \to 2} f(x) \) exists, the left-hand and right-hand limits must be equal:
\( 3 = 2 - a \)
\( a = 2 - 3 \)
\( a = -1 \)
In simple words: For the limit to exist at the boundary between two pieces, you must choose the parameter so that both formulas give the same limiting value as you approach that point.

Exam Tip: When asked to find an unknown parameter, set the left and right limits equal and solve - this is the core condition for limit existence at a boundary.

Question 8. Let \( f(x) = \begin{cases} \frac{3x}{|x| + 2x}, & x \neq 0 \\ 0, & x = 0 \end{cases} \) Show that \( \lim_{x \to 0} f(x) \) does not exist.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{3x}{|x| + 2x} = \lim_{x \to 0^-} \frac{3x}{(-x) + 2x} = \lim_{x \to 0^-} \frac{3x}{x} = \lim_{x \to 0^-} 3 = 3 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{3x}{|x| + 2x} = \lim_{x \to 0^+} \frac{3x}{(x) + 2x} = \lim_{x \to 0^+} \frac{3x}{3x} = \lim_{x \to 0^+} 1 = 1 \)

Since \( \lim_{x \to 0^-} f(x) = 3 \) and \( \lim_{x \to 0^+} f(x) = 1 \), the one-sided limits are unequal. Therefore, \( \lim_{x \to 0} f(x) \) does not exist.
In simple words: The absolute value in the denominator takes on different values depending on whether x is negative or positive, causing the entire fraction to simplify differently on each side.

Exam Tip: Always expand absolute value terms based on the sign of the variable in the region you are approaching from - this prevents sign errors.

Question 9. Let \( f(x) = \begin{cases} \cos x, & x \geq 0 \\ x + k, & x < 0 \end{cases} \) Find the value of k for which \( \lim_{x \to 0} f(x) \) exists.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + k) = 0 + k = k \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \cos x = \cos(0) = 1 \)

It is given that \( \lim_{x \to 0} f(x) \) exists. Therefore, the left-hand and right-hand limits must be equal:
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \)
\( k = 1 \)
In simple words: To make the limit exist, you need to pick k so that the linear piece on the left approaches the same value that the cosine piece approaches on the right as you converge to zero.

Exam Tip: For piecewise functions with a free parameter, equating the one-sided limits and solving directly gives you the required parameter value instantly.

Question 10. Show that \( \lim_{x \to 0} \frac{1}{x} \) does not exist.
Answer: To examine this limit, we use the substitution \( x = 0 + h \) as \( x \) approaches \( 0^+ \). Since \( x \to 0 \), we have \( h \to 0 \) as well.

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{x} = \lim_{h \to 0^+} \frac{1}{0 + h} = \lim_{h \to 0^+} \frac{1}{h} = \frac{1}{0} = +\infty \)

Now let \( x = 0 - h \) as \( x \) approaches \( 0^- \). Since \( x \to 0 \), we have \( h \to 0 \) as well.

Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{x} = \lim_{h \to 0^-} \frac{1}{0 - h} = \lim_{h \to 0^-} \frac{1}{-h} = \frac{1}{-0} = -\infty \)

Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), the limit does not exist.

Exam Tip: When examining limits at a point, always compute both the left and right hand limits separately. If these one-sided limits are not equal, the overall limit fails to exist at that point.

Question 11. Show that \( \lim_{x \to 0} \frac{1}{|x|} = \infty \).
Answer: We examine this limit by computing both the right and left hand limits. Let \( x = 0 + h \) as \( x \) tends to \( 0^+ \). Since \( x \to 0 \), we also have \( h \to 0 \).

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{|x|} = \lim_{x \to 0^+} \frac{1}{x} = +\infty \)

Now let \( x = 0 - h \) as \( x \) tends to \( 0^- \). Since \( x \to 0 \), we have \( h \to 0 \) as well.

Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{|x|} = \lim_{x \to 0^-} \frac{1}{-x} = \lim_{h \to 0^-} \frac{1}{h} = +\infty \)

Since both the left and right hand limits are equal and both approach \( +\infty \), we can conclude that \( \lim_{x \to 0} \frac{1}{|x|} = \infty \).

Exam Tip: When absolute value expressions are involved, carefully evaluate the function on each side of the limiting point. If both one-sided limits match and are infinite, the overall limit is infinite in that direction.

Question 12. Show that \( \lim_{x \to 0} e^{-1/x} \) does not exist.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{-1/x} = \lim_{x \to 0^-} e^{-1/(-x)} = \lim_{x \to 0^-} e^{1/x} = e^{\infty} = \infty \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{-1/x} = e^{-1/0} = e^{-\infty} = 0 \)

Since the left hand limit equals \( \infty \) and the right hand limit equals \( 0 \), we have \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \). Therefore, \( \lim_{x \to 0} e^{-1/x} \) does not exist.

Exam Tip: Exponential functions with negative exponents behave differently as you approach zero from the left versus the right. Always check whether the exponent itself approaches \( +\infty \) or \( -\infty \) from each direction.

Question 13. Show that \( \lim_{x \to 0} \sin \frac{1}{x} \) does not exist.
Answer: To verify this, we compute both one-sided limits. Let \( x = 0 + h \) when \( x \) tends to \( 0^+ \). Since \( x \to 0 \), we have \( h \to 0 \) as well.

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sin \frac{1}{x} = \lim_{h \to 0^+} \sin \frac{1}{0 + h} = \sin \frac{1}{0} = \sin \infty \)

The expression \( \sin \infty \) is undefined because sine oscillates between -1 and 1 indefinitely as its argument increases without bound.

Now let \( x = 0 - h \) when \( x \) tends to \( 0^- \). Since \( x \to 0 \), we have \( h \to 0 \) as well.

Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin \frac{1}{x} = \lim_{h \to 0^-} \sin \frac{1}{0 - h} = \sin \frac{1}{-0} = - \sin \frac{1}{0} = - \sin \infty \)

This is also undefined, and the oscillatory behavior differs from the right hand limit. Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), the limit does not exist.

Exam Tip: When trigonometric functions have arguments that approach infinity, the function oscillates wildly and no limit exists. Always recognize that \( \sin \infty \) and \( \cos \infty \) are undefined forms.

Question 14. Show that \( \lim_{x \to 0} \frac{x}{|x|} \) does not exist.
Answer:
Left Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{(-x)} = \lim_{x \to 0^-} (-1) = -1 \)

Right Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 = 1 \)

Since \( \lim_{x \to 0^-} f(x) = -1 \) and \( \lim_{x \to 0^+} f(x) = 1 \), and these are not equal, the limit does not exist.

Exam Tip: When working with absolute value expressions, remember that \( |x| = x \) for positive \( x \) and \( |x| = -x \) for negative \( x \). This causes the function to behave differently on each side of zero.

Question 15. Let \( f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases} \). If \( \lim_{x \to \pi/2} f(x) = f\left(\frac{\pi}{2}\right) \), find the value of k.
Answer: We need to find the limit of the function as \( x \) approaches \( \frac{\pi}{2} \):

\( \lim_{x \to \pi/2} f(x) = \lim_{x \to \pi/2} \frac{k \cos x}{\pi - 2x} \)

Let \( h = x - \frac{\pi}{2} \), so \( x = h + \frac{\pi}{2} \). As \( x \to \frac{\pi}{2} \), we have \( h \to 0 \).

Substituting, we get \( x \to \frac{\pi}{2} \), and since \( h + \frac{\pi}{2} \to \frac{\pi}{2} \), it follows that \( h \to 0 \).

\( = \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + h\right)}{\pi - 2\left(\frac{\pi}{2} + h\right)} = \lim_{h \to 0} \frac{-k \sin h}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \lim_{h \to 0} \frac{k \sin h}{2h} = -k \lim_{h \to 0} \frac{\sin h}{h} \)

Using the standard limit formula \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):

\( = -k \times 1 = -k \)

We are given that \( f\left(\frac{\pi}{2}\right) = 3 \).

For continuity at \( x = \frac{\pi}{2} \), we require \( \lim_{x \to \pi/2} f(x) = f\left(\frac{\pi}{2}\right) \):

\( -k = 3 \)

\( \therefore k = -3 \)

Exam Tip: When finding unknown parameters that ensure continuity, always apply the condition that the limit must equal the function value at that point. Use standard trigonometric limits like \( \frac{\sin x}{x} \to 1 \) to simplify indeterminate forms.