OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Exercise 9 (B)

Find by applying L'Hospital's Rule the following limits :

Question 1. \( \lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2} \)
Answer:
We need to evaluate the limit \( \lim_{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2} \).
First, substitute \( x=0 \) into the expression to check for the form:
Numerator: \( 0 \cdot e^0 - \log(1+0) = 0 - \log(1) = 0 - 0 = 0 \)
Denominator: \( 0^2 = 0 \)
So, this is of the \( \frac{0}{0} \) indeterminate form. We can apply L'Hospital's Rule.
L'Hospital's Rule states that if \( \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \) is of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \).
Differentiate the numerator \( f(x) = x e^x - \log(1+x) \):
\( f'(x) = \frac{d}{dx}(x e^x) - \frac{d}{dx}(\log(1+x)) \)
\( = (1 \cdot e^x + x \cdot e^x) - \frac{1}{1+x} \)
\( = e^x + x e^x - \frac{1}{1+x} \)
Differentiate the denominator \( g(x) = x^2 \):
\( g'(x) = 2x \)
Now, apply L'Hospital's Rule for the first time:
\( \lim_{x \rightarrow 0} \frac{e^x + x e^x - \frac{1}{1+x}}{2x} \)
Substitute \( x=0 \) again:
Numerator: \( e^0 + 0 \cdot e^0 - \frac{1}{1+0} = 1 + 0 - 1 = 0 \)
Denominator: \( 2 \cdot 0 = 0 \)
This is still of the \( \frac{0}{0} \) indeterminate form. So, we apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = e^x + x e^x - \frac{1}{1+x} \):
\( f_1'(x) = \frac{d}{dx}(e^x) + \frac{d}{dx}(x e^x) - \frac{d}{dx}((1+x)^{-1}) \)
\( = e^x + (1 \cdot e^x + x \cdot e^x) - (-1)(1+x)^{-2}(1) \)
\( = e^x + e^x + x e^x + \frac{1}{(1+x)^2} \)
\( = 2 e^x + x e^x + \frac{1}{(1+x)^2} \)
Differentiate the new denominator \( g_1(x) = 2x \):
\( g_1'(x) = 2 \)
Now, apply L'Hospital's Rule for the second time:
\( \lim_{x \rightarrow 0} \frac{2 e^x + x e^x + \frac{1}{(1+x)^2}}{2} \)
Substitute \( x=0 \):
\( = \frac{2 e^0 + 0 \cdot e^0 + \frac{1}{(1+0)^2}}{2} \)
\( = \frac{2(1) + 0 + \frac{1}{1}}{2} \)
\( = \frac{2+1}{2} \)
\( = \frac{3}{2} \)
In simple words: When you have a limit problem where plugging in the number gives you 0/0, you can use L'Hospital's Rule. This means you take the derivative of the top part and the bottom part separately, and then try the limit again. We had to do this twice here until we got a clear answer.

🎯 Exam Tip: Remember to check for the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hospital's Rule. Differentiate numerator and denominator separately, not as a quotient. Sometimes, you might need to apply the rule multiple times.

Question 2. \( \lim _{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3} \)
Answer:
We need to evaluate the limit \( \lim_{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3} \).
First, substitute \( x=0 \) into the expression:
Numerator: \( 0 - \sin(0) \cos(0) = 0 - 0 \cdot 1 = 0 \)
Denominator: \( 0^3 = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form, so we can apply L'Hospital's Rule.
We can simplify \( \sin x \cos x \) to \( \frac{1}{2} \sin(2x) \).
So, the expression becomes \( \lim_{x \rightarrow 0} \frac{x-\frac{1}{2} \sin(2x)}{x^3} \).
Differentiate the numerator \( f(x) = x - \frac{1}{2} \sin(2x) \):
\( f'(x) = 1 - \frac{1}{2} \cdot \cos(2x) \cdot 2 = 1 - \cos(2x) \)
Differentiate the denominator \( g(x) = x^3 \):
\( g'(x) = 3x^2 \)
Apply L'Hospital's Rule for the first time:
\( \lim_{x \rightarrow 0} \frac{1-\cos(2x)}{3x^2} \)
Substitute \( x=0 \) again:
Numerator: \( 1 - \cos(0) = 1 - 1 = 0 \)
Denominator: \( 3 \cdot 0^2 = 0 \)
This is still of the \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = 1 - \cos(2x) \):
\( f_1'(x) = 0 - (-\sin(2x)) \cdot 2 = 2 \sin(2x) \)
Differentiate the new denominator \( g_1(x) = 3x^2 \):
\( g_1'(x) = 6x \)
Apply L'Hospital's Rule for the second time:
\( \lim_{x \rightarrow 0} \frac{2 \sin(2x)}{6x} \)
This can be written as \( \frac{2}{6} \lim_{x \rightarrow 0} \frac{\sin(2x)}{x} = \frac{1}{3} \lim_{x \rightarrow 0} \frac{\sin(2x)}{x} \).
We know that \( \lim_{y \rightarrow 0} \frac{\sin y}{y} = 1 \). Let \( y = 2x \). As \( x \rightarrow 0 \), \( y \rightarrow 0 \).
So, \( \lim_{x \rightarrow 0} \frac{\sin(2x)}{x} = \lim_{x \rightarrow 0} \frac{2 \sin(2x)}{2x} = 2 \cdot 1 = 2 \).
Therefore, the limit is \( \frac{1}{3} \cdot 2 = \frac{2}{3} \).
Alternatively, we could apply L'Hospital's Rule one more time to \( \lim_{x \rightarrow 0} \frac{2 \sin(2x)}{6x} \).
Numerator derivative: \( \frac{d}{dx}(2 \sin(2x)) = 2 \cos(2x) \cdot 2 = 4 \cos(2x) \)
Denominator derivative: \( \frac{d}{dx}(6x) = 6 \)
So, \( \lim_{x \rightarrow 0} \frac{4 \cos(2x)}{6} = \frac{4 \cos(0)}{6} = \frac{4 \cdot 1}{6} = \frac{4}{6} = \frac{2}{3} \).
In simple words: This problem also started as 0/0, so we used L'Hospital's Rule more than once. We simplified the sine and cosine product first, then took derivatives of the top and bottom separately twice until we could directly find the limit.

🎯 Exam Tip: Simplifying trigonometric expressions using identities like \( \sin A \cos A = \frac{1}{2} \sin(2A) \) can often make differentiation easier and reduce calculation errors in L'Hospital's Rule problems.

Question 3. \( \lim _{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3} \).
First, substitute \( x=0 \):
Numerator: \( e^0 \sin(0) - 0 - 0^2 = 1 \cdot 0 - 0 - 0 = 0 \)
Denominator: \( 0^3 = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We must apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = e^x \sin x - x - x^2 \):
\( f'(x) = (e^x \sin x + e^x \cos x) - 1 - 2x = e^x(\sin x + \cos x) - 1 - 2x \)
Differentiate the denominator \( g(x) = x^3 \):
\( g'(x) = 3x^2 \)
Apply L'Hospital's Rule (1st time):
\( \lim_{x \rightarrow 0} \frac{e^x(\sin x + \cos x) - 1 - 2x}{3x^2} \)
Substitute \( x=0 \):
Numerator: \( e^0(\sin(0) + \cos(0)) - 1 - 2(0) = 1(0+1) - 1 - 0 = 1 - 1 = 0 \)
Denominator: \( 3(0)^2 = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = e^x(\sin x + \cos x) - 1 - 2x \):
\( f_1'(x) = (e^x(\sin x + \cos x) + e^x(\cos x - \sin x)) - 0 - 2 \)
\( = e^x(\sin x + \cos x + \cos x - \sin x) - 2 \)
\( = e^x(2 \cos x) - 2 = 2e^x \cos x - 2 \)
Differentiate the new denominator \( g_1(x) = 3x^2 \):
\( g_1'(x) = 6x \)
Apply L'Hospital's Rule (2nd time):
\( \lim_{x \rightarrow 0} \frac{2e^x \cos x - 2}{6x} \)
Substitute \( x=0 \):
Numerator: \( 2e^0 \cos(0) - 2 = 2(1)(1) - 2 = 2 - 2 = 0 \)
Denominator: \( 6(0) = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_2(x) = 2e^x \cos x - 2 \):
\( f_2'(x) = (2e^x \cos x + 2e^x (-\sin x)) - 0 = 2e^x(\cos x - \sin x) \)
Differentiate the new denominator \( g_2(x) = 6x \):
\( g_2'(x) = 6 \)
Apply L'Hospital's Rule (3rd time):
\( \lim_{x \rightarrow 0} \frac{2e^x(\cos x - \sin x)}{6} \)
Substitute \( x=0 \):
\( = \frac{2e^0(\cos(0) - \sin(0))}{6} \)
\( = \frac{2(1)(1 - 0)}{6} \)
\( = \frac{2 \cdot 1}{6} = \frac{2}{6} = \frac{1}{3} \)
In simple words: This problem required us to apply L'Hospital's Rule three times because each time we tried the limit, we still got the 0/0 form. We differentiated the top and bottom separately at each step until we could get a definite numerical answer.

🎯 Exam Tip: Be very careful with product rule and chain rule when differentiating complex functions repeatedly. Organize your derivatives clearly to avoid mistakes, especially when dealing with \( e^x \) and trigonometric functions.

Question 4. \( \lim _{x \rightarrow \pi / 4} \frac{\sin \left(x+\frac{\pi}{4}\right)-1}{\log \sin 2x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow \pi / 4} \frac{\sin \left(x+\frac{\pi}{4}\right)-1}{\log \sin 2x} \).
First, substitute \( x = \frac{\pi}{4} \):
Numerator: \( \sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) - 1 = \sin\left(\frac{2\pi}{4}\right) - 1 = \sin\left(\frac{\pi}{2}\right) - 1 = 1 - 1 = 0 \)
Denominator: \( \log\left(\sin\left(2 \cdot \frac{\pi}{4}\right)\right) = \log\left(\sin\left(\frac{\pi}{2}\right)\right) = \log(1) = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We can apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \sin\left(x+\frac{\pi}{4}\right)-1 \):
\( f'(x) = \cos\left(x+\frac{\pi}{4}\right) \cdot 1 - 0 = \cos\left(x+\frac{\pi}{4}\right) \)
Differentiate the denominator \( g(x) = \log(\sin 2x) \):
\( g'(x) = \frac{1}{\sin 2x} \cdot \cos 2x \cdot 2 = 2 \frac{\cos 2x}{\sin 2x} = 2 \cot 2x \)
Apply L'Hospital's Rule:
\( \lim_{x \rightarrow \pi / 4} \frac{\cos\left(x+\frac{\pi}{4}\right)}{2 \cot 2x} \)
Substitute \( x = \frac{\pi}{4} \) again:
Numerator: \( \cos\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \)
Denominator: \( 2 \cot\left(2 \cdot \frac{\pi}{4}\right) = 2 \cot\left(\frac{\pi}{2}\right) = 2 \cdot 0 = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = \cos\left(x+\frac{\pi}{4}\right) \):
\( f_1'(x) = -\sin\left(x+\frac{\pi}{4}\right) \)
Differentiate the new denominator \( g_1(x) = 2 \cot 2x \):
\( g_1'(x) = 2 \cdot (-\text{cosec}^2 2x) \cdot 2 = -4 \text{cosec}^2 2x \)
Apply L'Hospital's Rule for the second time:
\( \lim_{x \rightarrow \pi / 4} \frac{-\sin\left(x+\frac{\pi}{4}\right)}{-4 \text{cosec}^2 2x} = \lim_{x \rightarrow \pi / 4} \frac{\sin\left(x+\frac{\pi}{4}\right)}{4 \text{cosec}^2 2x} \)
Substitute \( x = \frac{\pi}{4} \):
\( = \frac{\sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right)}{4 \text{cosec}^2\left(2 \cdot \frac{\pi}{4}\right)} \)
\( = \frac{\sin\left(\frac{\pi}{2}\right)}{4 \text{cosec}^2\left(\frac{\pi}{2}\right)} \)
\( = \frac{1}{4 \cdot (1)^2} = \frac{1}{4} \)
The cosecant of \( \frac{\pi}{2} \) is 1, because \( \sin(\frac{\pi}{2}) = 1 \).
In simple words: After plugging in \( \pi/4 \), we got 0/0, so L'Hospital's Rule was needed. We differentiated the top and bottom functions twice. It's important to be careful with chain rules, especially for trigonometric and logarithmic functions.

🎯 Exam Tip: When evaluating limits involving trigonometric functions, carefully apply the chain rule during differentiation. Remember that \( \frac{d}{dx}(\log u) = \frac{u'}{u} \) and \( \frac{d}{dx}(\cot u) = - \text{cosec}^2 u \cdot u' \). Keep track of signs.

Question 5. \( \lim_{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^4} \)
Answer:
We need to evaluate the limit \( \lim_{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^4} \).
First, substitute \( x=0 \):
Numerator: \( e^0 + e^{-0} + 2 \cos(0) - 4 = 1 + 1 + 2(1) - 4 = 1+1+2-4 = 0 \)
Denominator: \( 0^4 = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We must apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = e^x+e^{-x}+2 \cos x-4 \):
\( f'(x) = e^x - e^{-x} - 2 \sin x \)
Differentiate the denominator \( g(x) = x^4 \):
\( g'(x) = 4x^3 \)
Apply L'Hospital's Rule (1st time):
\( \lim_{x \rightarrow 0} \frac{e^x - e^{-x} - 2 \sin x}{4x^3} \)
Substitute \( x=0 \):
Numerator: \( e^0 - e^{-0} - 2 \sin(0) = 1 - 1 - 0 = 0 \)
Denominator: \( 4(0)^3 = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = e^x - e^{-x} - 2 \sin x \):
\( f_1'(x) = e^x - (-e^{-x}) - 2 \cos x = e^x + e^{-x} - 2 \cos x \)
Differentiate the new denominator \( g_1(x) = 4x^3 \):
\( g_1'(x) = 12x^2 \)
Apply L'Hospital's Rule (2nd time):
\( \lim_{x \rightarrow 0} \frac{e^x + e^{-x} - 2 \cos x}{12x^2} \)
Substitute \( x=0 \):
Numerator: \( e^0 + e^{-0} - 2 \cos(0) = 1 + 1 - 2(1) = 0 \)
Denominator: \( 12(0)^2 = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_2(x) = e^x + e^{-x} - 2 \cos x \):
\( f_2'(x) = e^x - e^{-x} + 2 \sin x \)
Differentiate the new denominator \( g_2(x) = 12x^2 \):
\( g_2'(x) = 24x \)
Apply L'Hospital's Rule (3rd time):
\( \lim_{x \rightarrow 0} \frac{e^x - e^{-x} + 2 \sin x}{24x} \)
Substitute \( x=0 \):
Numerator: \( e^0 - e^{-0} + 2 \sin(0) = 1 - 1 + 0 = 0 \)
Denominator: \( 24(0) = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_3(x) = e^x - e^{-x} + 2 \sin x \):
\( f_3'(x) = e^x + e^{-x} + 2 \cos x \)
Differentiate the new denominator \( g_3(x) = 24x \):
\( g_3'(x) = 24 \)
Apply L'Hospital's Rule (4th time):
\( \lim_{x \rightarrow 0} \frac{e^x + e^{-x} + 2 \cos x}{24} \)
Substitute \( x=0 \):
\( = \frac{e^0 + e^{-0} + 2 \cos(0)}{24} \)
\( = \frac{1 + 1 + 2(1)}{24} \)
\( = \frac{4}{24} = \frac{1}{6} \)
In simple words: This problem was a bit longer as we had to use L'Hospital's Rule four times in a row. Each time, we differentiated the top and bottom of the fraction separately because plugging in zero kept giving us 0/0. Eventually, we reached a point where we could calculate the limit directly.

🎯 Exam Tip: For problems requiring multiple applications of L'Hospital's Rule, ensure each differentiation step is correct. Keep your work organized. A common pattern with \( e^x + e^{-x} \) is that its derivative is \( e^x - e^{-x} \), and the next derivative is back to \( e^x + e^{-x} \), which can help in checking your work.

Question 6. \( \lim _{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x} \).
First, substitute \( x=0 \):
Numerator: \( e^0 - e^{-0} - 2 \log(1+0) = 1 - 1 - 2 \log(1) = 0 - 0 = 0 \)
Denominator: \( 0 \cdot \sin(0) = 0 \cdot 0 = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We will apply L'Hospital's Rule.
We can rewrite the denominator \( x \sin x \). As \( x \rightarrow 0 \), \( \sin x \approx x \). So \( x \sin x \approx x^2 \).
This means we can also use the standard limit \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \).
The expression can be written as \( \lim_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x^2} \cdot \lim_{x \rightarrow 0} \frac{x}{\sin x} \).
Since \( \lim_{x \rightarrow 0} \frac{x}{\sin x} = 1 \), we only need to evaluate \( \lim_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x^2} \).
Differentiate the numerator \( f(x) = e^x-e^{-x}-2 \log (1+x) \):
\( f'(x) = e^x - (-e^{-x}) - 2 \cdot \frac{1}{1+x} = e^x + e^{-x} - \frac{2}{1+x} \)
Differentiate the denominator \( g(x) = x^2 \):
\( g'(x) = 2x \)
Apply L'Hospital's Rule (1st time):
\( \lim_{x \rightarrow 0} \frac{e^x + e^{-x} - \frac{2}{1+x}}{2x} \)
Substitute \( x=0 \):
Numerator: \( e^0 + e^{-0} - \frac{2}{1+0} = 1 + 1 - 2 = 0 \)
Denominator: \( 2(0) = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = e^x + e^{-x} - 2(1+x)^{-1} \):
\( f_1'(x) = e^x - e^{-x} - 2(-1)(1+x)^{-2}(1) = e^x - e^{-x} + \frac{2}{(1+x)^2} \)
Differentiate the new denominator \( g_1(x) = 2x \):
\( g_1'(x) = 2 \)
Apply L'Hospital's Rule (2nd time):
\( \lim_{x \rightarrow 0} \frac{e^x - e^{-x} + \frac{2}{(1+x)^2}}{2} \)
Substitute \( x=0 \):
\( = \frac{e^0 - e^{-0} + \frac{2}{(1+0)^2}}{2} \)
\( = \frac{1 - 1 + \frac{2}{1}}{2} \)
\( = \frac{0 + 2}{2} = \frac{2}{2} = 1 \)
So the value of the limit is 1.
In simple words: This limit problem also started with 0/0. We first used a known limit trick for \( \frac{x}{\sin x} \), which made the problem simpler. Then, we applied L'Hospital's Rule twice by differentiating the top and bottom of the remaining fraction separately, until we got our answer.

🎯 Exam Tip: Before blindly applying L'Hospital's Rule, consider if you can simplify the expression using standard limits like \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \) or \( \lim_{x \rightarrow 0} \frac{\tan x}{x} = 1 \). This can often reduce the number of differentiation steps needed.

Question 7. \( \lim _{x \rightarrow \frac{1}{2}} \frac{\cos ^2 \pi x}{e^{2 x}-2 e x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow \frac{1}{2}} \frac{\cos ^2 \pi x}{e^{2 x}-2 e x} \).
First, substitute \( x = \frac{1}{2} \):
Numerator: \( \cos^2\left(\pi \cdot \frac{1}{2}\right) = \cos^2\left(\frac{\pi}{2}\right) = 0^2 = 0 \)
Denominator: \( e^{2 \cdot \frac{1}{2}} - 2e \cdot \frac{1}{2} = e^1 - e = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We will apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \cos^2 \pi x \):
\( f'(x) = 2 \cos(\pi x) \cdot (-\sin(\pi x)) \cdot \pi = -2\pi \sin(\pi x) \cos(\pi x) \)
We can also use the identity \( 2 \sin A \cos A = \sin(2A) \):
\( f'(x) = -\pi \sin(2\pi x) \)
Differentiate the denominator \( g(x) = e^{2x} - 2ex \):
\( g'(x) = e^{2x} \cdot 2 - 2e = 2e^{2x} - 2e \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow \frac{1}{2}} \frac{-\pi \sin(2\pi x)}{2e^{2x} - 2e} \)
Substitute \( x = \frac{1}{2} \):
Numerator: \( -\pi \sin\left(2\pi \cdot \frac{1}{2}\right) = -\pi \sin(\pi) = -\pi \cdot 0 = 0 \)
Denominator: \( 2e^{2 \cdot \frac{1}{2}} - 2e = 2e^1 - 2e = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = -\pi \sin(2\pi x) \):
\( f_1'(x) = -\pi \cos(2\pi x) \cdot 2\pi = -2\pi^2 \cos(2\pi x) \)
Differentiate the new denominator \( g_1(x) = 2e^{2x} - 2e \):
\( g_1'(x) = 2e^{2x} \cdot 2 - 0 = 4e^{2x} \)
Apply L'Hospital's Rule (2nd time):
\( \lim _{x \rightarrow \frac{1}{2}} \frac{-2\pi^2 \cos(2\pi x)}{4e^{2x}} \)
Substitute \( x = \frac{1}{2} \):
\( = \frac{-2\pi^2 \cos\left(2\pi \cdot \frac{1}{2}\right)}{4e^{2 \cdot \frac{1}{2}}} \)
\( = \frac{-2\pi^2 \cos(\pi)}{4e^1} \)
\( = \frac{-2\pi^2 (-1)}{4e} \)
\( = \frac{2\pi^2}{4e} = \frac{\pi^2}{2e} \)
In simple words: When we put \( x=1/2 \) into the original limit, we got 0/0. So, we used L'Hospital's Rule twice. This means we took the derivative of the top and bottom separately, simplified, and repeated the process until we could find a clear answer for the limit.

🎯 Exam Tip: Recognize and use trigonometric identities like \( \sin(2A) = 2 \sin A \cos A \) to simplify derivatives, as this can make calculations more manageable. Be careful with constants like \( \pi \) and \( e \) during differentiation and evaluation.

Question 8. \( \lim _{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x} \).
First, substitute \( x=0 \):
Numerator: \( \log(1-0^2) = \log(1) = 0 \)
Denominator: \( \log(\cos(0)) = \log(1) = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We will apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \log(1-x^2) \):
\( f'(x) = \frac{1}{1-x^2} \cdot (-2x) = \frac{-2x}{1-x^2} \)
Differentiate the denominator \( g(x) = \log \cos x \):
\( g'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\frac{-2x}{1-x^2}}{-\tan x} = \lim _{x \rightarrow 0} \frac{2x}{(1-x^2)\tan x} \)
We can rewrite this limit as: \( \lim_{x \rightarrow 0} \frac{2}{1-x^2} \cdot \lim_{x \rightarrow 0} \frac{x}{\tan x} \)
We know that \( \lim_{x \rightarrow 0} \frac{\tan x}{x} = 1 \), so \( \lim_{x \rightarrow 0} \frac{x}{\tan x} = 1 \).
Now, evaluate each part of the limit:
\( \lim_{x \rightarrow 0} \frac{2}{1-x^2} = \frac{2}{1-0^2} = \frac{2}{1} = 2 \)
\( \lim_{x \rightarrow 0} \frac{x}{\tan x} = 1 \)
So, the overall limit is \( 2 \cdot 1 = 2 \).
In simple words: This problem involved limits of logarithmic functions. Since plugging in zero gave us 0/0, we used L'Hospital's Rule once. After differentiating, we separated the expression into two simpler limits. One of these was a standard limit involving \( \tan x \), which helped us solve the problem.

🎯 Exam Tip: When faced with products or quotients after applying L'Hospital's Rule, try to separate them into known standard limits or simpler parts before re-applying the rule. This can often lead to a quicker solution and fewer steps.

Question 9. \( \lim _{x \rightarrow 0} \frac{\tan x-x}{x^2 \tan x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\tan x-x}{x^2 \tan x} \).
First, substitute \( x=0 \):
Numerator: \( \tan(0) - 0 = 0 - 0 = 0 \)
Denominator: \( 0^2 \cdot \tan(0) = 0 \cdot 0 = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We will apply L'Hospital's Rule.
We can rewrite the expression as \( \lim _{x \rightarrow 0} \frac{\tan x-x}{x^3} \cdot \frac{x}{\tan x} \).
Since \( \lim _{x \rightarrow 0} \frac{x}{\tan x} = 1 \), we only need to evaluate \( \lim _{x \rightarrow 0} \frac{\tan x-x}{x^3} \).
Differentiate the numerator \( f(x) = \tan x-x \):
\( f'(x) = \sec^2 x - 1 \)
Differentiate the denominator \( g(x) = x^3 \):
\( g'(x) = 3x^2 \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\sec^2 x - 1}{3x^2} \)
We know that \( \sec^2 x - 1 = \tan^2 x \). So the expression becomes \( \lim _{x \rightarrow 0} \frac{\tan^2 x}{3x^2} \).
This can be written as \( \lim _{x \rightarrow 0} \frac{1}{3} \left(\frac{\tan x}{x}\right)^2 \).
Since \( \lim_{x \rightarrow 0} \frac{\tan x}{x} = 1 \), then \( \lim_{x \rightarrow 0} \frac{1}{3} \left(\frac{\tan x}{x}\right)^2 = \frac{1}{3} (1)^2 = \frac{1}{3} \).
Alternatively, if we continue with L'Hospital's Rule for \( \lim _{x \rightarrow 0} \frac{\sec^2 x - 1}{3x^2} \):
Substitute \( x=0 \):
Numerator: \( \sec^2(0) - 1 = 1 - 1 = 0 \)
Denominator: \( 3(0)^2 = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = \sec^2 x - 1 \):
\( f_1'(x) = 2 \sec x (\sec x \tan x) = 2 \sec^2 x \tan x \)
Differentiate the new denominator \( g_1(x) = 3x^2 \):
\( g_1'(x) = 6x \)
Apply L'Hospital's Rule (2nd time):
\( \lim _{x \rightarrow 0} \frac{2 \sec^2 x \tan x}{6x} \)
This can be written as \( \lim _{x \rightarrow 0} \frac{2 \sec^2 x}{6} \cdot \frac{\tan x}{x} = \frac{1}{3} \lim _{x \rightarrow 0} \sec^2 x \cdot \lim _{x \rightarrow 0} \frac{\tan x}{x} \).
Substitute \( x=0 \) into the remaining parts:
\( \frac{1}{3} \cdot \sec^2(0) \cdot 1 = \frac{1}{3} \cdot (1)^2 \cdot 1 = \frac{1}{3} \).
Both methods yield the same result.
In simple words: This problem was about limits with tangent functions. We first separated a known limit part \( \frac{x}{\tan x} \). For the remaining 0/0 form, we used L'Hospital's Rule twice. We found that \( \sec^2 x - 1 \) is the same as \( \tan^2 x \), which helped simplify the calculations.

🎯 Exam Tip: Using standard limits like \( \lim_{x \rightarrow 0} \frac{\tan x}{x} = 1 \) can greatly simplify problems, reducing the need for multiple L'Hospital's Rule applications. Also, remember key trigonometric identities like \( \sec^2 x - 1 = \tan^2 x \).

Question 10. \( \lim _{x \rightarrow 0} \frac{\sin x-\log \left(e^x \cos x\right)}{x \sin x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\sin x-\log \left(e^x \cos x\right)}{x \sin x} \).
First, substitute \( x=0 \):
Numerator: \( \sin(0) - \log(e^0 \cos(0)) = 0 - \log(1 \cdot 1) = 0 - \log(1) = 0 - 0 = 0 \)
Denominator: \( 0 \cdot \sin(0) = 0 \cdot 0 = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We will apply L'Hospital's Rule.
First, simplify the numerator using logarithm properties: \( \log(AB) = \log A + \log B \).
\( \log(e^x \cos x) = \log(e^x) + \log(\cos x) = x + \log(\cos x) \).
So the numerator becomes \( \sin x - (x + \log(\cos x)) = \sin x - x - \log(\cos x) \).
The expression is \( \lim _{x \rightarrow 0} \frac{\sin x - x - \log(\cos x)}{x \sin x} \).
As in previous problems, we can separate \( \frac{x}{\sin x} \) from the denominator, which goes to 1 as \( x \rightarrow 0 \).
So, we focus on \( \lim _{x \rightarrow 0} \frac{\sin x - x - \log(\cos x)}{x^2} \).
Differentiate the numerator \( f(x) = \sin x - x - \log(\cos x) \):
\( f'(x) = \cos x - 1 - \frac{1}{\cos x} (-\sin x) = \cos x - 1 + \frac{\sin x}{\cos x} = \cos x - 1 + \tan x \)
Differentiate the denominator \( g(x) = x^2 \):
\( g'(x) = 2x \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\cos x - 1 + \tan x}{2x} \)
Substitute \( x=0 \):
Numerator: \( \cos(0) - 1 + \tan(0) = 1 - 1 + 0 = 0 \)
Denominator: \( 2(0) = 0 \)
Still \( \frac{0}{0} \) indeterminate form. Apply L'Hospital's Rule again.
Differentiate the new numerator \( f_1(x) = \cos x - 1 + \tan x \):
\( f_1'(x) = -\sin x - 0 + \sec^2 x = -\sin x + \sec^2 x \)
Differentiate the new denominator \( g_1(x) = 2x \):
\( g_1'(x) = 2 \)
Apply L'Hospital's Rule (2nd time):
\( \lim _{x \rightarrow 0} \frac{-\sin x + \sec^2 x}{2} \)
Substitute \( x=0 \):
\( = \frac{-\sin(0) + \sec^2(0)}{2} \)
\( = \frac{0 + 1^2}{2} = \frac{1}{2} \)
In simple words: This limit problem began with 0/0. First, we simplified the logarithm in the numerator and used a standard limit for the denominator. Then, we applied L'Hospital's Rule twice, differentiating the top and bottom of the remaining fraction separately, until we could get a clear number.

🎯 Exam Tip: Always simplify logarithmic expressions using properties like \( \log(AB) = \log A + \log B \) before differentiating. This can simplify the expression greatly and make subsequent differentiation steps easier and less prone to errors.

Question 11. \( \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x} \).
First, substitute \( x=0 \):
Numerator: \( \log(\sin(0)) = \log(0) \rightarrow -\infty \)
Denominator: \( \cot(0) \rightarrow \infty \)
This is of the \( \frac{-\infty}{\infty} \) indeterminate form. We can apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \log \sin x \):
\( f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \)
Differentiate the denominator \( g(x) = \cot x \):
\( g'(x) = -\text{cosec}^2 x \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\cot x}{-\text{cosec}^2 x} \)
We can rewrite this expression using \( \cot x = \frac{\cos x}{\sin x} \) and \( \text{cosec}^2 x = \frac{1}{\sin^2 x} \):
\( = \lim _{x \rightarrow 0} \frac{\frac{\cos x}{\sin x}}{-\frac{1}{\sin^2 x}} \)
\( = \lim _{x \rightarrow 0} \frac{\cos x}{\sin x} \cdot (-\sin^2 x) \)
\( = \lim _{x \rightarrow 0} (-\cos x \sin x) \)
Substitute \( x=0 \):
\( = - \cos(0) \sin(0) = -1 \cdot 0 = 0 \)
In simple words: This limit problem was of the infinity/infinity form. We used L'Hospital's Rule once, differentiating the top and bottom parts. Then, we simplified the trigonometric terms to find the final limit.

🎯 Exam Tip: When using L'Hospital's Rule with trigonometric functions, converting all terms to \( \sin x \) and \( \cos x \) often helps simplify the expression after differentiation, making it easier to evaluate the limit directly.

Question 12. \( \lim _{x \rightarrow 0} \frac{\log \tan x}{\log x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\log \tan x}{\log x} \).
First, substitute \( x=0 \):
Numerator: \( \log(\tan(0)) = \log(0) \rightarrow -\infty \)
Denominator: \( \log(0) \rightarrow -\infty \)
This is of the \( \frac{-\infty}{-\infty} \) indeterminate form. We can apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \log \tan x \):
\( f'(x) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} \)
Differentiate the denominator \( g(x) = \log x \):
\( g'(x) = \frac{1}{x} \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x \cos x}}{\frac{1}{x}} = \lim _{x \rightarrow 0} \frac{x}{\sin x \cos x} \)
We can split this into known limits: \( \lim _{x \rightarrow 0} \frac{x}{\sin x} \cdot \lim _{x \rightarrow 0} \frac{1}{\cos x} \).
We know that \( \lim_{x \rightarrow 0} \frac{x}{\sin x} = 1 \).
And \( \lim_{x \rightarrow 0} \frac{1}{\cos x} = \frac{1}{\cos(0)} = \frac{1}{1} = 1 \).
So, the limit is \( 1 \cdot 1 = 1 \).
In simple words: This limit problem involved two logarithmic functions and gave us the infinity/infinity form. We used L'Hospital's Rule once by differentiating the top and bottom. Then, we reorganized the terms to use a known limit \( \frac{x}{\sin x} \) and directly substituted for the other part to find the answer.

🎯 Exam Tip: When dealing with logarithmic functions and trigonometric functions, it is often helpful to simplify the derivative expressions using fundamental identities before evaluating the limit. Also, look for opportunities to use standard limits like \( \lim_{x \rightarrow 0} \frac{x}{\sin x} = 1 \).

Question 13. \( \lim _{x \rightarrow 0} \frac{\log \tan 2 x}{\log \tan x} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\log \tan 2 x}{\log \tan x} \).
First, substitute \( x=0 \):
Numerator: \( \log(\tan(0)) = \log(0) \rightarrow -\infty \)
Denominator: \( \log(\tan(0)) = \log(0) \rightarrow -\infty \)
This is of the \( \frac{-\infty}{-\infty} \) indeterminate form. We will apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \log \tan 2x \):
\( f'(x) = \frac{1}{\tan 2x} \cdot \sec^2 2x \cdot 2 = \frac{2 \sec^2 2x}{\tan 2x} \)
Differentiate the denominator \( g(x) = \log \tan x \):
\( g'(x) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\frac{2 \sec^2 2x}{\tan 2x}}{\frac{\sec^2 x}{\tan x}} = \lim _{x \rightarrow 0} \frac{2 \sec^2 2x}{\tan 2x} \cdot \frac{\tan x}{\sec^2 x} \)
We can rearrange this as: \( 2 \cdot \lim _{x \rightarrow 0} \frac{\sec^2 2x}{\sec^2 x} \cdot \lim _{x \rightarrow 0} \frac{\tan x}{\tan 2x} \)
First part: \( \lim _{x \rightarrow 0} \frac{\sec^2 2x}{\sec^2 x} = \frac{\sec^2(0)}{\sec^2(0)} = \frac{1^2}{1^2} = 1 \).
Second part: \( \lim _{x \rightarrow 0} \frac{\tan x}{\tan 2x} \)
Substitute \( x=0 \): \( \frac{\tan(0)}{\tan(0)} = \frac{0}{0} \). This part is still indeterminate. We can apply L'Hospital's Rule to this sub-limit.
For \( \lim _{x \rightarrow 0} \frac{\tan x}{\tan 2x} \):
Numerator derivative: \( \sec^2 x \)
Denominator derivative: \( \sec^2 2x \cdot 2 = 2 \sec^2 2x \)
So, \( \lim _{x \rightarrow 0} \frac{\sec^2 x}{2 \sec^2 2x} = \frac{\sec^2(0)}{2 \sec^2(0)} = \frac{1}{2 \cdot 1} = \frac{1}{2} \).
Now, combine the results:
\( 2 \cdot (1) \cdot \left(\frac{1}{2}\right) = 1 \).
In simple words: This problem had two logarithms of tangent functions, leading to an infinity/infinity form. We applied L'Hospital's Rule once. After differentiating, we needed to simplify the resulting trigonometric fraction, which still had an indeterminate part. We solved that smaller limit separately and then combined our results.

🎯 Exam Tip: When a limit expression involves multiple factors, sometimes a sub-limit remains indeterminate. Break down the problem into smaller, manageable limits. Applying L'Hospital's Rule to these sub-limits individually can clarify the overall solution.

Question 14. \( \lim _{x \rightarrow 0} \log (1-x) \cot \frac{\pi x}{2} \)
Answer:
We need to evaluate the limit \( \lim _{x \rightarrow 0} \log (1-x) \cot \frac{\pi x}{2} \).
First, substitute \( x=0 \):
\( \log(1-0) \cdot \cot\left(\frac{\pi \cdot 0}{2}\right) = \log(1) \cdot \cot(0) = 0 \cdot \infty \)
This is of the \( 0 \cdot \infty \) indeterminate form. We need to rewrite it as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) to apply L'Hospital's Rule.
We can rewrite \( \cot\left(\frac{\pi x}{2}\right) \) as \( \frac{1}{\tan\left(\frac{\pi x}{2}\right)} \).
So the expression becomes \( \lim _{x \rightarrow 0} \frac{\log (1-x)}{\tan\left(\frac{\pi x}{2}\right)} \).
Now, substitute \( x=0 \):
Numerator: \( \log(1-0) = \log(1) = 0 \)
Denominator: \( \tan\left(\frac{\pi \cdot 0}{2}\right) = \tan(0) = 0 \)
This is of the \( \frac{0}{0} \) indeterminate form. We can apply L'Hospital's Rule.
Differentiate the numerator \( f(x) = \log(1-x) \):
\( f'(x) = \frac{1}{1-x} \cdot (-1) = \frac{-1}{1-x} \)
Differentiate the denominator \( g(x) = \tan\left(\frac{\pi x}{2}\right) \):
\( g'(x) = \sec^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2} \)
Apply L'Hospital's Rule (1st time):
\( \lim _{x \rightarrow 0} \frac{\frac{-1}{1-x}}{\frac{\pi}{2} \sec^2\left(\frac{\pi x}{2}\right)} \)
Substitute \( x=0 \):
\( = \frac{\frac{-1}{1-0}}{\frac{\pi}{2} \sec^2\left(\frac{\pi \cdot 0}{2}\right)} \)
\( = \frac{-1}{\frac{\pi}{2} \sec^2(0)} \)
\( = \frac{-1}{\frac{\pi}{2} \cdot 1^2} = \frac{-1}{\frac{\pi}{2}} = -\frac{2}{\pi} \)
In simple words: This problem started as a 0 times infinity indeterminate form. To use L'Hospital's Rule, we first changed the expression into a 0/0 form by rewriting the cotangent. Then, we differentiated the top and bottom of the new fraction to find the limit.

🎯 Exam Tip: When a limit is in the \( 0 \cdot \infty \) form, always convert it to either \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hospital's Rule. This is typically done by moving one of the factors to the denominator as its reciprocal, for example, \( f(x)g(x) = \frac{f(x)}{1/g(x)} \) or \( \frac{g(x)}{1/f(x)} \).

Question 17. \( \lim _{x \rightarrow 0}(\cot x)^{\sin 2 x} \)
Answer: Let the limit be \( y \).
So, \( y = \lim _{x \rightarrow 0}(\cot x)^{\sin 2 x} \) (This is of the form \( \infty^0 \)).
To solve this, we take the natural logarithm on both sides.
\( \log y = \lim _{x \rightarrow 0} \sin 2x \log (\cot x) \) (This is of the form \( 0 \cdot \infty \)).
We can rewrite this expression to use L'Hospital's Rule.
\( \log y = \lim _{x \rightarrow 0} \frac{\log (\cot x)}{\frac{1}{\sin 2 x}} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\log (\cot x)}{\operatorname{cosec} 2 x} \) (This is of the form \( \frac{\infty}{\infty} \)).
Now, apply L'Hospital's Rule by differentiating the numerator and the denominator.
Derivative of \( \log(\cot x) \) is \( \frac{1}{\cot x} (-\operatorname{cosec}^2 x) \).
Derivative of \( \operatorname{cosec} 2x \) is \( -\operatorname{cosec} 2x \cot 2x \cdot 2 \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\frac{1}{\cot x}(-\operatorname{cosec}^2 x)}{-2 \operatorname{cosec} 2 x \cot 2 x} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-\frac{\cos x}{\sin x} \cdot \frac{1}{\sin^2 x}}{-2 \frac{\cos 2x}{\sin^2 2x} \cdot \frac{1}{\sin 2x}} \) (Using \( \cot x = \frac{\cos x}{\sin x} \) and \( \operatorname{cosec} x = \frac{1}{\sin x} \))
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-\frac{1}{\sin x \cos x}}{\frac{-2 \cos 2x}{\sin^3 2x}} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{1}{\sin x \cos x} \cdot \frac{\sin^3 2x}{2 \cos 2x} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{1}{\sin x \cos x} \cdot \frac{(2 \sin x \cos x)^3}{2 \cos 2x} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{1}{\sin x \cos x} \cdot \frac{8 \sin^3 x \cos^3 x}{2 \cos 2x} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{4 \sin^2 x \cos^2 x}{\cos 2x} \)
Now, substitute \( x=0 \).
\( \implies \log y = \frac{4 (\sin 0)^2 (\cos 0)^2}{\cos (2 \cdot 0)} \)
\( \implies \log y = \frac{4 \cdot 0^2 \cdot 1^2}{1} \)
\( \implies \log y = 0 \)
If \( \log y = 0 \), then \( y = e^0 \).
\( \implies y = 1 \)
Therefore, \( \lim _{x \rightarrow 0}(\cot x)^{\sin 2 x} = 1 \). This type of limit often simplifies nicely to a power of e.
In simple words: When we have a limit like this where the base goes to infinity and the power goes to zero, we take a logarithm to change it into a form we can solve. After applying L'Hospital's Rule and simplifying, the final answer comes out to be 1.

🎯 Exam Tip: For indeterminate forms of the type \( \infty^0 \), \( 0^0 \), or \( 1^\infty \), always take the natural logarithm of the function first to convert it into a \( 0 \cdot \infty \) or \( \frac{0}{0} \) form, which can then be solved using L'Hospital's Rule.

Question 18. \( \lim _{x \rightarrow 0}(1+\sin x)^{\cot x} \)
Answer: Let the limit be \( y \).
So, \( y = \lim _{x \rightarrow 0}(1+\sin x)^{\cot x} \) (This is of the form \( 1^\infty \)).
To solve this, we take the natural logarithm on both sides.
\( \log y = \lim _{x \rightarrow 0} \cot x \log (1+\sin x) \) (This is of the form \( \infty \cdot 0 \)).
We can rewrite this expression to use L'Hospital's Rule.
\( \log y = \lim _{x \rightarrow 0} \frac{\log (1+\sin x)}{\frac{1}{\cot x}} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\log (1+\sin x)}{\tan x} \) (This is of the form \( \frac{0}{0} \)).
Now, apply L'Hospital's Rule by differentiating the numerator and the denominator.
Derivative of \( \log(1+\sin x) \) is \( \frac{1}{1+\sin x} (\cos x) \).
Derivative of \( \tan x \) is \( \sec^2 x \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\frac{\cos x}{1+\sin x}}{\sec^2 x} \)
Now, substitute \( x=0 \).
\( \implies \log y = \frac{\frac{\cos 0}{1+\sin 0}}{\sec^2 0} \)
\( \implies \log y = \frac{\frac{1}{1+0}}{1^2} \)
\( \implies \log y = \frac{1}{1} \)
\( \implies \log y = 1 \)
If \( \log y = 1 \), then \( y = e^1 \).
\( \implies y = e \)
Therefore, \( \lim _{x \rightarrow 0}(1+\sin x)^{\cot x} = e \). This method helps convert complex limit problems into simpler forms.
In simple words: When the limit is in the form of "1 to the power of infinity," we use logarithms. This changes the problem into a form where we can apply L'Hospital's Rule. After doing the math, the final answer comes out to be the number 'e'.

🎯 Exam Tip: Remember the special limit \( \lim_{x \to 0} (1+f(x))^{1/f(x)} = e \) when \( \lim_{x \to 0} f(x) = 0 \). This question is a variation of that type, where \( f(x) = \sin x \) and \( \cot x = 1/\tan x \approx 1/x \) for small x.

Question 19. \( \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 / x^2} \)
Answer: Let the limit be \( y \).
So, \( y = \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 / x^2} \) (This is of the form \( 1^\infty \), because \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{1}{x^2} = \infty \)).
To solve this, we take the natural logarithm on both sides.
\( \log y = \lim _{x \rightarrow 0} \frac{1}{x^2} \log \left(\frac{\sin x}{x}\right) \) (This is of the form \( \infty \cdot 0 \)).
We can rewrite this expression to use L'Hospital's Rule.
\( \log y = \lim _{x \rightarrow 0} \frac{\log \left(\frac{\sin x}{x}\right)}{x^2} \) (This is of the form \( \frac{0}{0} \)).
Now, apply L'Hospital's Rule by differentiating the numerator and the denominator.
Derivative of \( \log\left(\frac{\sin x}{x}\right) \) is \( \frac{1}{\frac{\sin x}{x}} \cdot \frac{x \cos x - \sin x}{x^2} = \frac{x}{\sin x} \cdot \frac{x \cos x - \sin x}{x^2} = \frac{x \cos x - \sin x}{x \sin x} \).
Derivative of \( x^2 \) is \( 2x \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\frac{x \cos x - \sin x}{x \sin x}}{2x} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{x \cos x - \sin x}{2x^2 \sin x} \)
This is still of the form \( \frac{0}{0} \). Apply L'Hospital's Rule again.
Derivative of \( x \cos x - \sin x \) is \( (1 \cdot \cos x + x(-\sin x)) - \cos x = \cos x - x \sin x - \cos x = -x \sin x \).
Derivative of \( 2x^2 \sin x \) is \( 2(2x \sin x + x^2 \cos x) = 4x \sin x + 2x^2 \cos x \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-x \sin x}{4x \sin x + 2x^2 \cos x} \)
Divide the numerator and denominator by \( x \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-\sin x}{4 \sin x + 2x \cos x} \)
This is still of the form \( \frac{0}{0} \). Apply L'Hospital's Rule again.
Derivative of \( -\sin x \) is \( -\cos x \).
Derivative of \( 4 \sin x + 2x \cos x \) is \( 4 \cos x + (2 \cdot \cos x + 2x(-\sin x)) = 4 \cos x + 2 \cos x - 2x \sin x = 6 \cos x - 2x \sin x \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-\cos x}{6 \cos x - 2x \sin x} \)
Now, substitute \( x=0 \).
\( \implies \log y = \frac{-\cos 0}{6 \cos 0 - 2(0) \sin 0} \)
\( \implies \log y = \frac{-1}{6 \cdot 1 - 0} \)
\( \implies \log y = -\frac{1}{6} \)
If \( \log y = -\frac{1}{6} \), then \( y = e^{-1/6} \).
Therefore, \( \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 / x^2} = e^{-1/6} \). Repeated application of L'Hospital's rule is often needed for such complex limits.
In simple words: For this type of limit, where the base goes to 1 and the power goes to infinity, we use logarithms. This changes it into a form where we can use L'Hospital's Rule multiple times. After several steps of differentiation, the final answer becomes \( e \) raised to the power of negative one-sixth.

🎯 Exam Tip: When L'Hospital's Rule still yields an indeterminate form, keep applying it until you reach a determinate value. Be careful with differentiation, especially product rules and chain rules.

Question 20. \( \lim _{x \rightarrow 0}\left(\frac{e^x-e^{-x}-2 x}{x-\sin x}\right) \)
Answer: Let the limit be \( L \).
\( L = \lim _{x \rightarrow 0}\left(\frac{e^x-e^{-x}-2 x}{x-\sin x}\right) \)
Substitute \( x=0 \): \( \frac{e^0-e^{-0}-2(0)}{0-\sin 0} = \frac{1-1-0}{0-0} = \frac{0}{0} \) (This is of the form \( \frac{0}{0} \)).
Apply L'Hospital's Rule by differentiating the numerator and the denominator.
Derivative of \( e^x-e^{-x}-2x \) is \( e^x - (-e^{-x}) - 2 = e^x + e^{-x} - 2 \).
Derivative of \( x-\sin x \) is \( 1-\cos x \).
\( L = \lim _{x \rightarrow 0}\left(\frac{e^x+e^{-x}-2}{1-\cos x}\right) \)
Substitute \( x=0 \): \( \frac{e^0+e^{-0}-2}{1-\cos 0} = \frac{1+1-2}{1-1} = \frac{0}{0} \) (This is still of the form \( \frac{0}{0} \)).
Apply L'Hospital's Rule again.
Derivative of \( e^x+e^{-x}-2 \) is \( e^x - e^{-x} \).
Derivative of \( 1-\cos x \) is \( \sin x \).
\( L = \lim _{x \rightarrow 0}\left(\frac{e^x-e^{-x}}{\sin x}\right) \)
Substitute \( x=0 \): \( \frac{e^0-e^{-0}}{\sin 0} = \frac{1-1}{0} = \frac{0}{0} \) (This is still of the form \( \frac{0}{0} \)).
Apply L'Hospital's Rule one more time.
Derivative of \( e^x-e^{-x} \) is \( e^x + e^{-x} \).
Derivative of \( \sin x \) is \( \cos x \).
\( L = \lim _{x \rightarrow 0}\left(\frac{e^x+e^{-x}}{\cos x}\right) \)
Now, substitute \( x=0 \).
\( L = \frac{e^0+e^{-0}}{\cos 0} \)
\( L = \frac{1+1}{1} \)
\( L = 2 \)
Therefore, \( \lim _{x \rightarrow 0}\left(\frac{e^x-e^{-x}-2 x}{x-\sin x}\right) = 2 \). Always perform repeated differentiation carefully until the indeterminate form is resolved.
In simple words: This problem involves an indeterminate form of 0/0. We use L'Hospital's Rule three times in a row. Each time, we differentiate the top and bottom parts of the fraction. After the third time, we can put \( x=0 \) into the expression to get the final answer, which is 2.

🎯 Exam Tip: When dealing with indeterminate forms, remember that L'Hospital's Rule can be applied multiple times. Keep differentiating until you can substitute the limit value without getting an indeterminate form.