ML Aggarwal Class 12 Maths Solutions Section A Chapter 07 Indeterminate Forms

Introduction

When the limits of both f(x) and g(x) exist as x approaches c, and the limit of g(x) is not zero, we can divide the limits: \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} \). But what happens when the limit of g(x) is zero? In such a case, for the limit of the fraction to exist (as a finite number), the limit of f(x) must also be zero.

To see why, suppose the limit of the fraction equals some value l. Then the limit of f(x) equals the limit of the fraction times the limit of g(x), which is l times 0, giving 0.

When both \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \), the fraction \( \frac{f(x)}{g(x)} \) is said to take the indeterminate form \( \frac{0}{0} \) as x approaches c. Other indeterminate forms that arise include \( \frac{\infty}{\infty} \), \( \infty - \infty \), \( 0 \cdot \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \).

Section 7.1 - Indeterminate Form \( \frac{0}{0} \)

L'Hôpital's Rule

If f(x) and g(x) are differentiable, g'(x) is not zero for all x in the interval (c - δ, c + δ) except possibly at x = c, \( \lim_{x \to c} f(x) = 0 = \lim_{x \to c} g(x) \), and the limit of \( \frac{f'(x)}{g'(x)} \) exists (either as a finite number or as infinity), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

(We accept this without proof.)

Remark: L'Hôpital's rule remains valid when the limit as x approaches c is replaced by one-sided limits.

Example 1. Evaluate the following limits:

(i) \( \lim_{x \to 4} \frac{x^4 - 256}{x^2 - 16} \)
When we substitute x = 4, we get the \( \frac{0}{0} \) form. Using L'Hôpital's rule, differentiate the numerator and denominator:
\( = \lim_{x \to 4} \frac{4x^3}{2x} = \lim_{x \to 4} 2x^2 = 2(4)^2 = 32 \)
In simple words: When both the top and bottom become zero at x = 4, we take derivatives of each and then substitute 4 to get 32.

(ii) \( \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} \)
At x = π/2, we get the \( \frac{0}{0} \) form. Applying L'Hôpital's rule:
\( = \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-1} = \lim_{x \to \frac{\pi}{2}} \sin x = \sin \frac{\pi}{2} = 1 \)
In simple words: Taking derivatives of top and bottom and then substituting gives us 1.

(iii) \( \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{\cos 2x} \)
Substituting x = π/4 gives the \( \frac{0}{0} \) form. Using L'Hôpital's rule:
\( = \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{-2\sin 2x} = \lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x}{2\sin 2x} \)
\( = \frac{\sec^2 \frac{\pi}{4}}{2\sin \frac{\pi}{2}} = \frac{(\sqrt{2})^2}{2 \cdot 1} = \frac{2}{2} = 1 \)
In simple words: Differentiate top and bottom separately, then substitute π/4 to arrive at the answer 1.

Example 2. Evaluate the following limits:

(i) \( \lim_{x \to 0} \frac{\sin x - x + \frac{1}{6}x^3}{x^3} \)
At x = 0, both numerator and denominator approach 0, so we have the \( \frac{0}{0} \) form. Applying L'Hôpital's rule three times:
First application: \( = \lim_{x \to 0} \frac{\cos x - 1 + \frac{1}{2}x^2}{3x^2} \) (still \( \frac{0}{0} \))
Second application: \( = \lim_{x \to 0} \frac{-\sin x + x}{6x} \) (still \( \frac{0}{0} \))
Third application: \( = \lim_{x \to 0} \frac{-\cos x + 1}{6} = \frac{-1 + 1}{6} = 0 \)
In simple words: We use L'Hôpital's rule repeatedly - three times total - taking derivatives at each step until the \( \frac{0}{0} \) form disappears.

(ii) \( \lim_{x \to 0} \frac{xe^x - \log(1 + x)}{x^2} \)
At x = 0, we get \( \frac{0}{0} \). Using L'Hôpital's rule twice:
First application: \( = \lim_{x \to 0} \frac{e^x + xe^x - \frac{1}{1+x}}{2x} \) (still \( \frac{0}{0} \))
Second application: \( = \lim_{x \to 0} \frac{e^x + e^x + xe^x + \frac{1}{(1+x)^2}}{2} = \frac{1 + 1 + 0 + 1}{2} = \frac{3}{2} \)
In simple words: Apply L'Hôpital's rule twice, differentiating both numerator and denominator each time, then substitute x = 0 to get 3/2.

Example 3. Evaluate the following limits:

(i) \( \lim_{x \to 1} \frac{1 + \log x - x}{1 - 2x + x^2} \)
At x = 1, we get the \( \frac{0}{0} \) form. Using L'Hôpital's rule:
\( = \lim_{x \to 1} \frac{0 + \frac{1}{x} - 1}{0 - 2 + 2x} = \lim_{x \to 1} \frac{\frac{1}{x} - 1}{-2 + 2x} = \lim_{x \to 1} \frac{\frac{1-x}{x}}{2(x-1)} \)
\( = \lim_{x \to 1} \frac{1 - x}{2x(x-1)} = \lim_{x \to 1} \frac{-1}{2x} = -\frac{1}{2} \)
In simple words: Differentiate the top and bottom, then substitute x = 1 to get negative one-half.

(ii) \( \lim_{x \to 0} \frac{e^x + e^{-x} + 2\cos x - 4}{x^4} \)
At x = 0, we have the \( \frac{0}{0} \) form. We apply L'Hôpital's rule four times:
First: \( = \lim_{x \to 0} \frac{e^x - e^{-x} - 2\sin x}{4x^3} \)
Second: \( = \lim_{x \to 0} \frac{e^x + e^{-x} - 2\cos x}{12x^2} \)
Third: \( = \lim_{x \to 0} \frac{e^x - e^{-x} + 2\sin x}{24x} \)
Fourth: \( = \lim_{x \to 0} \frac{e^x + e^{-x} + 2\cos x}{24} = \frac{1 + 1 + 2}{24} = \frac{1}{6} \)
In simple words: Keep taking derivatives of top and bottom four times in total until we can substitute x = 0 and get 1/6.

Example 4. Evaluate the following limits:

(i) \( \lim_{x \to \pi^-} \frac{\sin x}{\sqrt{x - \pi}} \)
At x = π, we have the \( \frac{0}{0} \) form. Applying L'Hôpital's rule:
\( = \lim_{x \to \pi^-} \frac{\cos x}{\frac{1}{2\sqrt{x - \pi}}} = \lim_{x \to \pi^-} 2\sqrt{x - \pi} \cos x = 0 \)
Note: The expression \( \sqrt{x - \pi} \) is not defined for x less than π, so the left-hand limit does not exist.
In simple words: Use L'Hôpital's rule to differentiate top and bottom, then as x approaches π from the right, the answer is 0.

(ii) \( \lim_{x \to 0} \frac{\log(1 + x^3)}{\sin^3 x} \)
We can rewrite this as: \( \lim_{x \to 0} \frac{\log(1 + x^3)}{x^3} \cdot \left(\frac{x}{\sin x}\right)^3 \)
Since \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \), we need: \( \lim_{x \to 0} \frac{\log(1 + x^3)}{x^3} = \lim_{x \to 0} \frac{\log(1 + x^3)}{x^3} \cdot 1^3 \)
Applying L'Hôpital's rule: \( = \lim_{x \to 0} \frac{\frac{1}{1+x^3} \cdot 3x^2}{3x^2} = \lim_{x \to 0} \frac{1}{1 + x^3} = 1 \)
In simple words: Rewrite the limit as a product, use the fact that x over sin x approaches 1, then apply L'Hôpital's rule to find the answer is 1.

Example 5. Evaluate the following limits:

(i) \( \lim_{x \to 0} \frac{x - \tan^{-1} x}{x - \sin x} \)
At x = 0, we get the \( \frac{0}{0} \) form. Applying L'Hôpital's rule:
\( = \lim_{x \to 0} \frac{1 - \frac{1}{1+x^2}}{1 - \cos x} \) (still \( \frac{0}{0} \))
\( = \lim_{x \to 0} \frac{\frac{1+x^2-1}{1+x^2}}{\sin x} = \lim_{x \to 0} \frac{x^2}{(1+x^2)\sin x} \)
\( = \lim_{x \to 0} \frac{2x}{2x\sin x + (1+x^2)\cos x} = \lim_{x \to 0} \frac{2}{2\sin x + 2x\cos x + 2x\cos x - (1+x^2)\sin x} \)
After simplifying: \( = 2 \)
In simple words: Apply L'Hôpital's rule, simplify carefully by combining fractions and differentiating again as needed, then substitute x = 0 to get 2.

(ii) \( \lim_{x \to 0} \frac{1 + \sin x - \cos x + \log(1 - x)}{x\tan^2 x} \)
At x = 0, we have the \( \frac{0}{0} \) form. After applying L'Hôpital's rule multiple times and simplifying through each application:
Following careful differentiation and substitution, the final answer is \( -\frac{1}{2} \)
In simple words: Apply L'Hôpital's rule several times, each time simplifying the numerator and denominator by taking derivatives until we reach a form where substitution gives us negative one-half.

Example 6. What is the fallacy in the following use of L'Hôpital's rule?

\( \lim_{x \to 1} \frac{x^3 + 3x - 4}{2x^2 + x - 3} = \lim_{x \to 1} \frac{3x^2 + 3}{4x + 1} = \lim_{x \to 1} \frac{6x}{4} = \frac{3}{2} \)

Solution: The error lies in applying L'Hôpital's rule when it does not apply. When we substitute x = 1 into the function \( \frac{3x^2 + 3}{4x + 1} \), we get \( \frac{3(1)^2 + 3}{4(1) + 1} = \frac{6}{5} \), which is not of the form \( \frac{0}{0} \). Therefore, L'Hôpital's rule cannot be used. The correct limit is found by direct substitution or by factoring and canceling the common factor, giving \( \frac{6}{5} \).

In simple words: L'Hôpital's rule only works when both top and bottom equal zero. Here the second fraction does not satisfy that, so we cannot use the rule again.

Exercise 7.1

Evaluate the following limits (1 to 13):

Question 1. (i) \( \lim_{x \to 3} \frac{x^4 - 81}{x - 3} \)

Answer: Using L'Hôpital's rule or factoring, we find the limit is 108.
In simple words: When x = 3, the numerator \( 3^4 - 81 = 0 \) and the denominator is 0, so we apply L'Hôpital's rule or factor the numerator to get 108.

Question 1. (ii) \( \lim_{x \to 0} \frac{(1+x)^n - 1}{x} \)

Answer: The limit equals n.
In simple words: Using L'Hôpital's rule, we differentiate to get n times the derivative of (1+x)^(n-1), which at x = 0 gives n.

Question 2. (i) \( \lim_{x \to 0} \frac{\sin ax}{\sin bx} \)

Answer: The limit is \( \frac{a}{b} \).
In simple words: Applying L'Hôpital's rule: differentiate numerator to get a cos(ax) and denominator to get b cos(bx). At x = 0, this gives a/b.

Question 2. (ii) \( \lim_{x \to 2} \frac{e^x - e^2}{x - 2} \)

Answer: The limit is \( e^2 \).
In simple words: By L'Hôpital's rule, the derivative of the numerator is \( e^x \) and of the denominator is 1. At x = 2, we get \( e^2 \).

Question 3. (i) \( \lim_{x \to 0} \frac{e^x - 1 - e^x}{1 - e^x} \)

Answer: The limit is -1.
In simple words: This simplifies to \( \lim_{x \to 0} \frac{-1}{1 - e^x} \), and as x approaches 0, the denominator approaches 0, which gives -1.

Question 3. (ii) \( \lim_{x \to 0} \frac{e^x - 1}{\tan 2x} \)

Answer: The limit is \( \frac{1}{2} \).
In simple words: Using L'Hôpital's rule, the numerator derivative is \( e^x \) and the denominator derivative is \( 2\sec^2 2x \). At x = 0, we get 1/2.

Question 4. (i) \( \lim_{x \to 0} \frac{e^x - (1 + x)}{x^2} \)

Answer: The limit is \( \frac{1}{2} \).
In simple words: Apply L'Hôpital's rule twice. First: \( \frac{e^x - 1}{2x} \). Second: \( \frac{e^x}{2} = \frac{1}{2} \) at x = 0.

Question 4. (ii) \( \lim_{x \to 1} \frac{x^2 - x \log x - \log x - 1}{x - 1} \)

Answer: The limit is 2.
In simple words: Use L'Hôpital's rule by differentiating the numerator and denominator, then substitute x = 1 to get 2.

Question 5. (i) \( \lim_{x \to 0} \frac{\cos x - 1}{\cos 2x - 1} \)

Answer: The limit is \( \frac{1}{4} \).
In simple words: Apply L'Hôpital's rule: numerator derivative is \( -\sin x \), denominator derivative is \( -2\sin 2x \). Use \( \sin 2x = 2\sin x \cos x \) and substitute x = 0 to get 1/4.

Question 5. (ii) \( \lim_{x \to 0} \frac{8^x - 2^x}{4^x} \)

Answer: The limit is \( \frac{1}{2}\log 2 \).
In simple words: Rewrite using exponentials and apply L'Hôpital's rule to find the final answer.

Question 6. (i) \( \lim_{x \to 0} \frac{\tan x - \sin x}{x - \sin x} \)

Answer: The limit is -2.
In simple words: Apply L'Hôpital's rule multiple times. Numerator derivative: \( \sec^2 x - \cos x \); denominator derivative: \( 1 - \cos x \). Continue applying the rule until substitution gives -2.

Question 6. (ii) \( \lim_{x \to 0} \frac{x - \tan^{-1} x}{x - \sin x} \)

Answer: The limit is 1.
In simple words: Use L'Hôpital's rule, being careful with derivatives of inverse trig functions and trigonometric expressions, to obtain the answer 1.

Question 7. (i) \( \lim_{x \to 0} \frac{\log \sec 2x}{\log \sec x} \)

Answer: The limit is 4.
In simple words: Apply L'Hôpital's rule. The derivative of \( \log \sec \theta \) is \( \tan \theta \). So numerator becomes \( 2\tan 2x \) and denominator becomes \( \tan x \). Use \( \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \) and simplify.

Question 7. (ii) \( \lim_{x \to 0} \frac{2\cos 2x - 2\cos x}{\sin^2 x} \)

Answer: The limit is \( -\frac{3}{2} \).
In simple words: Apply L'Hôpital's rule twice. Each time, take derivatives of the numerator and denominator, and eventually substitute x = 0 to reach \( -\frac{3}{2} \).

Question 8. (i) \( \lim_{x \to 0} \frac{x - \sin x}{x^3} \)

Answer: The limit is \( \frac{1}{6} \).
In simple words: Apply L'Hôpital's rule three times. First: \( \frac{1 - \cos x}{3x^2} \). Second: \( \frac{\sin x}{6x} \). Third: \( \frac{\cos x}{6} = \frac{1}{6} \).

Question 8. (ii) \( \lim_{x \to 0} \frac{e^x - e^{-x} - 2 - \log(1 + x)}{\sin x} \)

Answer: The limit is 1.
In simple words: Use L'Hôpital's rule, carefully computing the derivatives of each term in the numerator and denominator, then substitute x = 0.

Question 9. (i) \( \lim_{x \to 0} \frac{e^x + e^{-x} + 2\cos x - 4}{x^3} \)

Answer: The limit is 0.
In simple words: Apply L'Hôpital's rule. The numerator becomes \( e^x - e^{-x} - 2\sin x \) and denominator becomes \( 3x^2 \). Continue applying until you get 0 after substituting x = 0.

Question 9. (ii) \( \lim_{x \to 0} \frac{2 - 2\cos x + \sin^2 x}{x^3} \)

Answer: The limit is \( \frac{1}{12} \).
In simple words: Apply L'Hôpital's rule multiple times, carefully computing derivatives at each stage until substituting x = 0 yields \( \frac{1}{12} \).

Question 10. (i) \( \lim_{x \to 0} \frac{\log(1 - x)}{x \tan \frac{\pi x}{2}} \)

Answer: The limit is \( -\frac{2}{\pi} \).
In simple words: Apply L'Hôpital's rule. The numerator derivative is \( \frac{-1}{1-x} \) and the denominator requires the product rule. After simplification, substituting x = 0 gives \( -\frac{2}{\pi} \).

Question 10. (ii) \( \lim_{x \to 0} \frac{1 - 2^{\log(1 + x)}}{\log(1 + x)} \)

Answer: The limit is 1.
In simple words: Use L'Hôpital's rule carefully, noting that the exponential expression requires the chain rule for differentiation.

Question 11. (i) \( \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} \)

Answer: The limit is 2.
In simple words: Apply L'Hôpital's rule multiple times. Each differentiation simplifies the fraction. Eventually, when you substitute x = 0, the answer is 2.

Question 11. (ii) \( \lim_{x \to 0} \frac{e^x - e^{-\sin x}}{\sin x} \)

Answer: The limit is 1.
In simple words: Use L'Hôpital's rule. For the numerator derivative, the second term requires careful application of the chain rule with the negative sine exponent.

Question 12. (i) \( \lim_{x \to 0} \frac{\log(1 - x^2)}{\log \cos x} \)

Answer: The limit is 2.
In simple words: Apply L'Hôpital's rule. The numerator derivative is \( \frac{-2x}{1-x^2} \) and the denominator derivative is \( \frac{-\tan x}{1} = -\tan x \). After simplification and substitution, you get 2.

Question 12. (ii) \( \lim_{x \to 0} \frac{e^x \sin x - x + x^2}{x^3} \)

Answer: The limit is \( \frac{1}{3} \).
In simple words: Apply L'Hôpital's rule three times. Use the product rule for the \( e^x \sin x \) term in each differentiation until you can substitute x = 0.

Question 13. (i) \( \lim_{x \to 0} \frac{3^x - 2^x}{\sqrt{x}} \)

Answer: The limit is 0.
In simple words: The numerator approaches 1 - 1 = 0 while the denominator approaches 0. Use L'Hôpital's rule carefully to find the answer is 0.

Question 13. (ii) \( \lim_{x \to 0} \frac{(1 + x)^n - nx - 1}{x^2}, n > 1 \)

Answer: The limit is \( \frac{n(n-1)}{2} \).
In simple words: Apply L'Hôpital's rule twice. First: \( \frac{n(1+x)^{n-1} - n}{2x} \). Second: \( \frac{n(n-1)(1+x)^{n-2}}{2} \). Substitute x = 0 to get \( \frac{n(n-1)}{2} \).

Question 14. What is the fallacy in the following use of L'Hôpital's rule?

Answer: \( \lim_{x \to 2} \frac{3x^2 - 2x - 1}{3x^2 - 6x + 3} \) is shown being repeatedly differentiated to reach a conclusion. The fallacy is that after one or two applications of L'Hôpital's rule, the resulting expression is no longer of the indeterminate form \( \frac{0}{0} \), so continuing to apply the rule gives incorrect results. The denominator \( 3x^2 - 6x + 3 = 3(x - 1)^2 \), which is zero at x = 1 (not x = 2), so the original setup itself is flawed. One must verify that both numerator and denominator are zero at the point before applying L'Hôpital's rule, and must stop applying the rule as soon as the indeterminate form is resolved.
In simple words: L'Hôpital's rule only works when both top and bottom are zero at the same point. Check this before using the rule, and stop using it once the form is no longer 0/0.

Section 7.2 - Indeterminate Form \( \frac{\infty}{\infty} \)

L'Hôpital's Rule

If f(x) and g(x) are differentiable, g'(x) is not zero for all x in the interval (c - δ, c + δ) except possibly at x = c, and \( \lim_{x \to c} f(x) \to \infty \), \( \lim_{x \to c} g(x) \to \infty \), and the limit of \( \frac{f'(x)}{g'(x)} \) exists (either finitely or infinitely), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

(We accept this without proof.)

Remark: L'Hôpital's rule also applies when x approaches negative infinity.

Section 7.2.1 - Indeterminate Forms \( \infty - \infty \) and \( 0 \cdot \infty \)

These forms can be handled by first transforming them into either \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form, then applying L'Hôpital's rule.

Example 1. Evaluate the following limits:

(i) \( \lim_{x \to \frac{\pi}{2}} \frac{\tan x}{\tan 3x} \)
At x = π/2, both the numerator and denominator approach infinity, giving the \( \frac{\infty}{\infty} \) form. Using L'Hôpital's rule:
\( = \lim_{x \to \frac{\pi}{2}} \frac{\sec^2 x}{\sec^2 3x \cdot 3} = \lim_{x \to \frac{\pi}{2}} \frac{\cos^2 3x}{3\cos^2 x} \)
After evaluating at x = π/2, we simplify to get 3.
In simple words: As x approaches π/2, both tan x and tan 3x go to infinity. Differentiate top and bottom, simplify using cos^2 identities, and the answer becomes 3.

(ii) \( \lim_{x \to \infty} \frac{e^x + 3x^3}{4e^x + 2x^2} \)
As x approaches infinity, both numerator and denominator approach infinity, so we have the \( \frac{\infty}{\infty} \) form. Using L'Hôpital's rule repeatedly:
First: \( = \lim_{x \to \infty} \frac{e^x + 9x^2}{4e^x + 4x} \)
Second: \( = \lim_{x \to \infty} \frac{e^x + 18x}{4e^x + 4} \)
Third: \( = \lim_{x \to \infty} \frac{e^x + 18}{4e^x} \)
Fourth: \( = \lim_{x \to \infty} \frac{e^x}{4e^x} = \frac{1}{4} \)
In simple words: Keep applying L'Hôpital's rule (differentiating top and bottom) until the exponential term dominates and we can evaluate the limit as 1/4.

Example 2. Evaluate the following limits:

(i) \( \lim_{x \to c^+} \frac{\log(x - c)}{\log(e^x - e^c)} \)
As x approaches c from the right, both log terms approach negative infinity. We have the \( \frac{\infty}{\infty} \) form. Using L'Hôpital's rule:
\( = \lim_{x \to c^+} \frac{\frac{1}{x-c}}{\frac{1}{e^x - e^c} \cdot e^x} = \lim_{x \to c^+} \frac{e^x - e^c}{(x - c)e^x} \)
Now we have the \( \frac{0}{0} \) form. Applying L'Hôpital's rule again:
\( = \lim_{x \to c^+} \frac{e^x}{e^x + (x - c)e^x} = \lim_{x \to c^+} \frac{1}{1 + (x - c)} = 1 \)
In simple words: When we get infinity/infinity from logarithms, take derivatives. This transforms to 0/0, so apply L'Hôpital's again to find the answer is 1.

(ii) \( \lim_{x \to 0^+} \log_{\sin 2x} \sin x \)
This can be rewritten as \( \lim_{x \to 0^+} \frac{\log \sin x}{\log \sin 2x} \), which is of the form \( \frac{\infty}{\infty} \). Using L'Hôpital's rule:
\( = \lim_{x \to 0^+} \frac{\frac{\cos x}{\sin x}}{\frac{2\cos 2x}{\sin 2x}} = \lim_{x \to 0^+} \frac{\cos x \sin 2x}{2\cos 2x \sin x} \)
Using \( \sin 2x = 2\sin x \cos x \) and simplifying:
\( = \lim_{x \to 0^+} \frac{\cos x \cdot 2\sin x \cos x}{2\cos 2x \sin x} = \lim_{x \to 0^+} \frac{\cos^2 x}{\cos 2x} = 1 \)
In simple words: Rewrite the logarithmic expression as a fraction, apply L'Hôpital's rule to get cotangent ratios, use double-angle identities to simplify, and the answer is 1.

Example 3. Evaluate the following limits:

(i) \( \lim_{x \to 1} \left( \frac{x}{x - 1} - \frac{1}{\log x} \right) \)
This is an indeterminate form \( \infty - \infty \). Combine the fractions over a common denominator:
\( = \lim_{x \to 1} \frac{x\log x - (x - 1)}{(x - 1)\log x} \)
Both numerator and denominator approach 0, giving the \( \frac{0}{0} \) form. Using L'Hôpital's rule twice, the final answer is \( \frac{1}{2} \).
In simple words: When you have "infinity minus infinity", combine the fractions into one fraction, then apply L'Hôpital's rule to the resulting fraction.

(ii) \( \lim_{x \to 0} \left( \csc x - \frac{1}{x} \right) \)
Rewriting: \( \lim_{x \to 0} \frac{x - \sin x}{x \sin x} \), which is of the form \( \frac{0}{0} \). Applying L'Hôpital's rule repeatedly leads to the answer 0.
In simple words: Combine the two terms into a single fraction. Then use L'Hôpital's rule to find the limit is 0.

(iii) \( \lim_{x \to 0} \left( \frac{2}{x^2} - \cot x \right) \cdot x \)
Rearranging: \( \lim_{x \to 0} \left( \frac{2\cos x - x}{x^2 \sin x} \right) \), which becomes a \( \frac{0}{0} \) form. After applying L'Hôpital's rule multiple times, the answer is \( -\frac{1}{3} \).
In simple words: Multiply through and combine terms into one fraction. The resulting form \( \frac{0}{0} \) allows us to apply L'Hôpital's rule to get \( -\frac{1}{3} \).

(iv) \( \lim_{x \to \frac{\pi}{2}} \left( \pi - 2x \right) \left( \tan x - \sec x \right) \)
As x approaches π/2, \( \pi - 2x \) approaches 0 while both tan x and sec x approach infinity. This is a \( 0 \cdot \infty \) form. After combining and applying L'Hôpital's rule, the final answer is \( -\frac{\pi}{2} \).
In simple words: Rewrite the product as a single fraction by moving one factor to the denominator. This transforms the \( 0 \cdot \infty \) form into \( \frac{0}{0} \), then use L'Hôpital's rule.

Exercise 7.2

Evaluate the following limits (1 to 15):

Question 1. (i) \( \lim_{x \to 3} \frac{x^4 - 81}{x - 3} \)

Answer: The limit is 0.
In simple words: This is a \( \frac{\infty}{\infty} \) form as x approaches 3. Apply L'Hôpital's rule to get 0.

Question 1. (ii) \( \lim_{x \to -\infty} \frac{(1 + x)^n - 1}{x} \)

Answer: The limit is \( -\infty \).
In simple words: As x approaches negative infinity, we have the \( \frac{\infty}{\infty} \) form. L'Hôpital's rule gives \( -\infty \).

Question 2. (i) \( \lim_{x \to 0} \frac{\sin ax}{\sin bx} \)

Answer: The limit is 0.
In simple words: Both sine terms approach 0 as x approaches 0. Use L'Hôpital's rule to find the limit is 0.

Question 2. (ii) \( \lim_{x \to 2} \frac{e^x - e^2}{x - 2} \)

Answer: The limit is 0.
In simple words: Apply L'Hôpital's rule to this \( \frac{0}{0} \) form to get 0.