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

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

Question 1. \( \lim _{x \rightarrow 2} \frac{x^3-8}{x^2-4} \)
Answer: First, we check the form of the limit by substituting \( x=2 \). We get \( \frac{2^3-8}{2^2-4} = \frac{8-8}{4-4} = \frac{0}{0} \), which is an indeterminate form.
Since it is in \( \frac{0}{0} \) form, we can apply L'Hospital's Rule. This rule says we can take the derivative of the numerator and the denominator separately.
\( \lim _{x \rightarrow 2} \frac{x^3-8}{x^2-4} \)
\( = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}(x^3-8)}{\frac{d}{dx}(x^2-4)} \)
\( = \lim _{x \rightarrow 2} \frac{3x^2}{2x} \)
Now, we substitute \( x=2 \) into the new expression:
\( = \frac{3(2)^2}{2(2)} = \frac{3 \times 4}{4} = \frac{12}{4} = 3 \)
Therefore, the limit of the function as \( x \) approaches 2 is 3.
In simple words: When you plug in the number and get \( \frac{0}{0} \), you can take the derivative of the top and bottom parts. Then, plug in the number again to find the final answer.

🎯 Exam Tip: Always verify the indeterminate form (\( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)) before applying L'Hospital's Rule; otherwise, your answer will be incorrect.

Question 2. \( \lim _{x \rightarrow 3} \frac{x^3-x^2-18}{x-3} \)
Answer: We substitute \( x=3 \) into the expression to check its form.
\( \frac{3^3-3^2-18}{3-3} = \frac{27-9-18}{0} = \frac{18-18}{0} = \frac{0}{0} \), which is an indeterminate form.
Since it is in \( \frac{0}{0} \) form, we apply L'Hospital's Rule by taking the derivative of the numerator and the denominator.
\( \lim _{x \rightarrow 3} \frac{x^3-x^2-18}{x-3} \)
\( = \lim _{x \rightarrow 3} \frac{\frac{d}{dx}(x^3-x^2-18)}{\frac{d}{dx}(x-3)} \)
\( = \lim _{x \rightarrow 3} \frac{3x^2-2x}{1} \)
Now, we substitute \( x=3 \) into the new expression:
\( = 3(3)^2 - 2(3) = 3 \times 9 - 6 = 27 - 6 = 21 \)
The function simplifies nicely after differentiation, making the limit easy to calculate.
In simple words: If you get \( \frac{0}{0} \) after putting the number into the equation, find the derivative of the top part and the bottom part. Then, put the number in again to get your answer.

🎯 Exam Tip: Remember that the derivative of a constant (like -18 or -3) is zero. This is crucial for correctly simplifying the expressions using L'Hospital's Rule.

Question 3. \( \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
\( \frac{\tan^{-1}(0)}{0} = \frac{0}{0} \), which is an indeterminate form.
As it's in \( \frac{0}{0} \) form, we can apply L'Hospital's Rule. We differentiate the numerator and the denominator.
\( \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\tan^{-1} x)}{\frac{d}{dx}(x)} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x^2}}{1} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{\frac{1}{1+0^2}}{1} = \frac{\frac{1}{1}}{1} = 1 \)
This limit is a standard result that often appears in calculus, demonstrating the usefulness of L'Hospital's Rule.
In simple words: When the limit is \( \frac{0}{0} \), take the derivative of the top and bottom. Then, put the number back in to get the final value.

🎯 Exam Tip: Be sure to recall the standard derivative for inverse trigonometric functions, especially \( \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2} \), as it's frequently tested.

Question 4. \( \lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x} \)
Answer: First, we evaluate the limit at \( x=0 \):
\( \frac{e^{\sin 0}-1}{0} = \frac{e^0-1}{0} = \frac{1-1}{0} = \frac{0}{0} \), which is an indeterminate form.
Since it's in \( \frac{0}{0} \) form, we apply L'Hospital's Rule by differentiating the numerator and the denominator.
\( \lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(e^{\sin x}-1)}{\frac{d}{dx}(x)} \)
\( = \lim _{x \rightarrow 0} \frac{e^{\sin x} \cos x}{1} \)
Now, we substitute \( x=0 \) into the new expression:
\( = e^{\sin 0} \cos 0 = e^0 \times 1 = 1 \times 1 = 1 \)
This problem combines the chain rule with L'Hospital's Rule effectively.
In simple words: If you get \( \frac{0}{0} \) from the limit, differentiate the top and bottom parts. Then, substitute the number back into the new expression to find the answer.

🎯 Exam Tip: Remember the chain rule for derivatives, especially when you have a function like \( e^{g(x)} \), where its derivative is \( e^{g(x)} \cdot g'(x) \).

Question 5. \( \lim_{x \rightarrow 1} \frac{x-1}{2 x^2-7 x+5} \)
Answer: Let's check the form of the limit by substituting \( x=1 \).
Numerator: \( 1-1 = 0 \)
Denominator: \( 2(1)^2 - 7(1) + 5 = 2 - 7 + 5 = -5 + 5 = 0 \)
So, the limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator.
\( \lim_{x \rightarrow 1} \frac{x-1}{2 x^2-7 x+5} \)
\( = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}(x-1)}{\frac{d}{dx}(2x^2-7x+5)} \)
\( = \lim_{x \rightarrow 1} \frac{1}{4x-7} \)
Now, we substitute \( x=1 \) into the new expression:
\( = \frac{1}{4(1)-7} = \frac{1}{4-7} = \frac{1}{-3} = -\frac{1}{3} \)
Factoring the denominator would also work here, showing two paths to the solution.
In simple words: When the limit gives \( \frac{0}{0} \), take the derivative of the top and bottom. Then, put the value of \( x \) back into the new equation to get the answer.

🎯 Exam Tip: Be careful with the signs when calculating derivatives, especially with negative terms like \( -7x \). A small error in a sign can lead to a completely different answer.

Question 6. \( \lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1} \)
Answer: Let's substitute \( x=0 \) into the expression:
Numerator: \( (1+0)^5-1 = 1^5-1 = 1-1 = 0 \)
Denominator: \( (1+0)^3-1 = 1^3-1 = 1-1 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator using the power rule \( \frac{d}{dx}(u^n) = nu^{n-1} \frac{du}{dx} \).
\( \lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}((1+x)^5-1)}{\frac{d}{dx}((1+x)^3-1)} \)
\( = \lim _{x \rightarrow 0} \frac{5(1+x)^{5-1} \times 1}{3(1+x)^{3-1} \times 1} \)
\( = \lim _{x \rightarrow 0} \frac{5(1+x)^4}{3(1+x)^2} \)
Now, we substitute \( x=0 \) into the simplified expression:
\( = \frac{5(1+0)^4}{3(1+0)^2} = \frac{5(1)^4}{3(1)^2} = \frac{5 \times 1}{3 \times 1} = \frac{5}{3} \)
The power rule for derivatives is especially useful in problems like this.
In simple words: When you get \( \frac{0}{0} \) for a limit, use the derivative rule for powers on the top and bottom parts. Then, substitute the number for \( x \) to find the final answer.

🎯 Exam Tip: Remember the power rule for differentiation: if \( y = u^n \), then \( \frac{dy}{dx} = n u^{n-1} \frac{du}{dx} \). For \( (1+x)^n \), \( \frac{du}{dx} \) is just 1.

Question 7. \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 x-\pi}{\cos x} \)
Answer: Let's check the form of the limit by substituting \( x = \frac{\pi}{2} \).
Numerator: \( 2(\frac{\pi}{2})-\pi = \pi-\pi = 0 \)
Denominator: \( \cos(\frac{\pi}{2}) = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator.
\( \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 x-\pi}{\cos x} \)
\( = \lim_{x \rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx}(2x-\pi)}{\frac{d}{dx}(\cos x)} \)
\( = \lim_{x \rightarrow \frac{\pi}{2}} \frac{2}{-\sin x} \)
Now, we substitute \( x = \frac{\pi}{2} \) into the new expression:
\( = \frac{2}{-\sin(\frac{\pi}{2})} = \frac{2}{-1} = -2 \)
This demonstrates L'Hospital's rule for trigonometric functions, which is common in calculus.
In simple words: If you get \( \frac{0}{0} \) when putting the value into the limit, take the derivative of the top and bottom. Then, put the value for \( x \) back into the new equation to get your answer.

🎯 Exam Tip: Keep your trigonometric derivatives handy: \( \frac{d}{dx}(\cos x) = -\sin x \) and \( \frac{d}{dx}(\sin x) = \cos x \). Also, know the values of sine and cosine for common angles like \( \frac{\pi}{2} \).

Question 8. \( \lim _{x \rightarrow a} \frac{\sin x-\sin \alpha}{x-\alpha} \)
Answer: Let's check the form of the limit by substituting \( x=\alpha \).
Numerator: \( \sin \alpha - \sin \alpha = 0 \)
Denominator: \( \alpha - \alpha = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \). Remember that \( \alpha \) is treated as a constant.
\( \lim _{x \rightarrow a} \frac{\sin x-\sin \alpha}{x-\alpha} \)
\( = \lim _{x \rightarrow a} \frac{\frac{d}{dx}(\sin x-\sin \alpha)}{\frac{d}{dx}(x-\alpha)} \)
\( = \lim _{x \rightarrow a} \frac{\cos x - 0}{1 - 0} \)
\( = \lim _{x \rightarrow a} \cos x \)
Now, we substitute \( x=\alpha \) into the new expression:
\( = \cos \alpha \)
This limit is actually the definition of the derivative of \( \sin x \) at \( x=\alpha \).
In simple words: When you have \( \frac{0}{0} \) for a limit, take the derivative of the top and bottom parts. Treat \( \alpha \) like a fixed number. Then, put \( \alpha \) back in for \( x \) to find the answer.

🎯 Exam Tip: Recognize that this specific limit is a direct application of the definition of a derivative. If \( f(x) = \sin x \), then \( f'(\alpha) = \lim_{x \to \alpha} \frac{f(x)-f(\alpha)}{x-\alpha} \).

Question 9. \( \lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( a^0-b^0 = 1-1 = 0 \)
Denominator: \( e^0-1 = 1-1 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator. Remember that \( \frac{d}{dx}(c^x) = c^x \log c \).
\( \lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(a^x-b^x)}{\frac{d}{dx}(e^x-1)} \)
\( = \lim _{x \rightarrow 0} \frac{a^x \log a - b^x \log b}{e^x} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{a^0 \log a - b^0 \log b}{e^0} = \frac{1 \cdot \log a - 1 \cdot \log b}{1} \)
\( = \log a - \log b \)
Using logarithm properties, this simplifies further to:
\( = \log_e \left(\frac{a}{b}\right) \)
The derivative rule for exponential functions is key here.
In simple words: If the limit gives \( \frac{0}{0} \), take the derivative of the top and bottom. Remember that the derivative of \( c^x \) is \( c^x \log c \). Then, put \( x=0 \) back into the new expression to find the answer.

🎯 Exam Tip: Familiarize yourself with derivatives of exponential functions: \( \frac{d}{dx}(e^x) = e^x \) and \( \frac{d}{dx}(c^x) = c^x \ln c \). Also, know the logarithm property \( \log A - \log B = \log (\frac{A}{B}) \).

Question 10. \( \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}} \)
Answer: Let's check the form of the limit by substituting \( \alpha = \frac{\pi}{4} \).
Numerator: \( \sin(\frac{\pi}{4})-\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} = 0 \)
Denominator: \( \frac{\pi}{4}-\frac{\pi}{4} = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( \alpha \).
\( \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}} \)
\( = \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\frac{d}{d\alpha}(\sin \alpha-\cos \alpha)}{\frac{d}{d\alpha}(\alpha-\frac{\pi}{4})} \)
\( = \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\cos \alpha - (-\sin \alpha)}{1-0} \)
\( = \lim _{\alpha \rightarrow \frac{\pi}{4}} (\cos \alpha + \sin \alpha) \)
Now, we substitute \( \alpha = \frac{\pi}{4} \) into the new expression:
\( = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \)
This problem is another example of how L'Hospital's rule simplifies trigonometric limits.
In simple words: If you get \( \frac{0}{0} \) for a limit with angles, take the derivative of the top and bottom parts. Then, put the angle value back in for \( \alpha \) to find the answer. Remember the values of sine and cosine for common angles.

🎯 Exam Tip: Be mindful of the independent variable; here it's \( \alpha \), so you differentiate with respect to \( \alpha \). The derivative of a constant term like \( -\frac{\pi}{4} \) is zero.

Question 11. \( \lim _{x \rightarrow a} \frac{(x+2)^{5 / 3}-(a+2)^{5 / 3}}{x-a} \)
Answer: Let's check the form of the limit by substituting \( x=a \).
Numerator: \( (a+2)^{5/3}-(a+2)^{5/3} = 0 \)
Denominator: \( a-a = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \). Treat \( (a+2)^{5/3} \) as a constant since it doesn't contain \( x \).
\( \lim _{x \rightarrow a} \frac{(x+2)^{5 / 3}-(a+2)^{5 / 3}}{x-a} \)
\( = \lim _{x \rightarrow a} \frac{\frac{d}{dx}((x+2)^{5/3}-(a+2)^{5/3})}{\frac{d}{dx}(x-a)} \)
\( = \lim _{x \rightarrow a} \frac{\frac{5}{3}(x+2)^{\frac{5}{3}-1} \times 1 - 0}{1-0} \)
\( = \lim _{x \rightarrow a} \frac{5}{3}(x+2)^{2/3} \)
Now, we substitute \( x=a \) into the new expression:
\( = \frac{5}{3}(a+2)^{2/3} \)
This problem also closely relates to the definition of a derivative, specifically for the function \( f(y) = y^{5/3} \).
In simple words: When you get \( \frac{0}{0} \) in a limit, take the derivative of the top and bottom parts. Remember to use the power rule for derivatives. Then, put \( a \) back in for \( x \) to get the final answer.

🎯 Exam Tip: This problem is a classic example of the definition of the derivative. If \( f(x) = (x+2)^{5/3} \), the limit is \( f'(a) \). Recognize the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \).

Question 12. \( \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( \sqrt{1+\sin 0}-\sqrt{1-\sin 0} = \sqrt{1+0}-\sqrt{1-0} = 1-1 = 0 \)
Denominator: \( 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator. Remember that \( \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx} \).
\( \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\sqrt{1+\sin x}-\sqrt{1-\sin x})}{\frac{d}{dx}(x)} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{1}{2\sqrt{1+\sin x}}(\cos x) - \frac{1}{2\sqrt{1-\sin x}}(-\cos x)}{1} \)
\( = \lim _{x \rightarrow 0} \left(\frac{\cos x}{2\sqrt{1+\sin x}} + \frac{\cos x}{2\sqrt{1-\sin x}}\right) \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{\cos 0}{2\sqrt{1+\sin 0}} + \frac{\cos 0}{2\sqrt{1-\sin 0}} \)
\( = \frac{1}{2\sqrt{1+0}} + \frac{1}{2\sqrt{1-0}} = \frac{1}{2\sqrt{1}} + \frac{1}{2\sqrt{1}} \)
\( = \frac{1}{2} + \frac{1}{2} = 1 \)
This problem involves the chain rule for square roots and trigonometric functions, which makes it a good test of derivative knowledge.
In simple words: When the limit gives \( \frac{0}{0} \), take the derivative of the top and bottom. Remember how to differentiate square roots and sine functions. Then, put \( x=0 \) back into the new equation to find the answer.

🎯 Exam Tip: Be very careful with the chain rule when differentiating terms like \( \sqrt{1-\sin x} \). The derivative of the inner function \( (1-\sin x) \) is \( -\cos x \), which introduces a crucial sign change.

Question 13. If \( f(1) = 1, f'(1) = 2 \), then find \( \lim _{x \rightarrow 1} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1} \)
Answer: Let's check the form of the limit by substituting \( x=1 \).
Numerator: \( \sqrt{f(1)}-1 = \sqrt{1}-1 = 1-1 = 0 \)
Denominator: \( \sqrt{1}-1 = 1-1 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \).
\( \lim _{x \rightarrow 1} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1} \)
\( = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(\sqrt{f(x)}-1)}{\frac{d}{dx}(\sqrt{x}-1)} \)
\( = \lim _{x \rightarrow 1} \frac{\frac{1}{2\sqrt{f(x)}} f'(x)}{\frac{1}{2\sqrt{x}}} \)
We can simplify this expression by multiplying the numerator and denominator by \( 2 \):
\( = \lim _{x \rightarrow 1} \frac{\sqrt{x} f'(x)}{\sqrt{f(x)}} \)
Now, we substitute \( x=1 \) into the new expression and use the given values \( f(1)=1 \) and \( f'(1)=2 \):
\( = \frac{\sqrt{1} \cdot f'(1)}{\sqrt{f(1)}} = \frac{1 \cdot 2}{\sqrt{1}} = \frac{2}{1} = 2 \)
This problem shows how L'Hospital's rule can be applied even when a function is given generally.
In simple words: When the limit gives \( \frac{0}{0} \), take the derivative of the top and bottom. Use the chain rule for \( \sqrt{f(x)} \). Then, put \( x=1 \) into the new equation, using the given values of \( f(1) \) and \( f'(1) \), to find the answer.

🎯 Exam Tip: Remember to use the chain rule when differentiating composite functions like \( \sqrt{f(x)} \), which yields \( \frac{1}{2\sqrt{f(x)}} f'(x) \). Incorrectly differentiating \( \sqrt{f(x)} \) as just \( \frac{1}{2\sqrt{f(x)}} \) is a common mistake.

Question 14. \( \lim_{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( 2^0-1 = 1-1 = 0 \)
Denominator: \( \sqrt{1+0}-1 = \sqrt{1}-1 = 1-1 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator. Remember that \( \frac{d}{dx}(c^x) = c^x \log c \) and \( \frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}} \).
\( \lim_{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1} \)
\( = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(2^x-1)}{\frac{d}{dx}(\sqrt{1+x}-1)} \)
\( = \lim_{x \rightarrow 0} \frac{2^x \log 2}{\frac{1}{2\sqrt{1+x}}} \)
We can rewrite this expression:
\( = \lim_{x \rightarrow 0} 2\sqrt{1+x} \cdot 2^x \log 2 \)
Now, we substitute \( x=0 \) into the new expression:
\( = 2\sqrt{1+0} \cdot 2^0 \log 2 \)
\( = 2\sqrt{1} \cdot 1 \cdot \log 2 = 2 \cdot 1 \cdot \log 2 = 2 \log 2 \)
This problem combines derivatives of exponential functions and square roots.
In simple words: If you get \( \frac{0}{0} \) for a limit, take the derivative of the top and bottom parts. Remember the rules for differentiating numbers raised to \( x \) and square roots. Then, put \( x=0 \) back into the new equation to find the answer.

🎯 Exam Tip: Don't forget that \( \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx} \). Also, recall \( \frac{d}{dx}(a^x) = a^x \ln a \). Practicing these standard derivatives is key.

Question 15. \( \lim _{x \rightarrow 0} \frac{(1+x)^{1 / 2}-(1-x)^{1 / 2}}{x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( (1+0)^{1/2}-(1-0)^{1/2} = 1^{1/2}-1^{1/2} = 1-1 = 0 \)
Denominator: \( 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \). Use the power rule \( \frac{d}{dx}(u^n) = nu^{n-1} \frac{du}{dx} \).
\( \lim _{x \rightarrow 0} \frac{(1+x)^{1 / 2}-(1-x)^{1 / 2}}{x} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}((1+x)^{1/2}-(1-x)^{1/2})}{\frac{d}{dx}(x)} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{1}{2}(1+x)^{\frac{1}{2}-1}(1) - \frac{1}{2}(1-x)^{\frac{1}{2}-1}(-1)}{1} \)
\( = \lim _{x \rightarrow 0} \left(\frac{1}{2}(1+x)^{-1/2} + \frac{1}{2}(1-x)^{-1/2}\right) \)
\( = \lim _{x \rightarrow 0} \left(\frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}}\right) \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{1}{2\sqrt{1+0}} + \frac{1}{2\sqrt{1-0}} \)
\( = \frac{1}{2\sqrt{1}} + \frac{1}{2\sqrt{1}} = \frac{1}{2} + \frac{1}{2} = 1 \)
This result is a common limit identity that can be proven using binomial expansion or L'Hospital's rule.
In simple words: When the limit gives \( \frac{0}{0} \), use the power rule and chain rule to differentiate the top and bottom. Then, put \( x=0 \) back into the new equation to find the answer.

🎯 Exam Tip: Be careful with the chain rule for the \( (1-x)^{1/2} \) term. The derivative of \( (1-x) \) is \( -1 \), which leads to a positive sign in the simplified expression.

Question 16. \( \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( (1+0)^n-1 = 1^n-1 = 1-1 = 0 \)
Denominator: \( 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \). Use the power rule for \( (1+x)^n \).
\( \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}((1+x)^n-1)}{\frac{d}{dx}(x)} \)
\( = \lim _{x \rightarrow 0} \frac{n(1+x)^{n-1} \times 1 - 0}{1} \)
\( = \lim _{x \rightarrow 0} n(1+x)^{n-1} \)
Now, we substitute \( x=0 \) into the new expression:
\( = n(1+0)^{n-1} = n(1)^{n-1} = n \times 1 = n \)
This is a fundamental limit that is also the definition of the derivative for \( x^n \) at \( x=1 \), or more generally, for \( (1+x)^n \) at \( x=0 \).
In simple words: When the limit gives \( \frac{0}{0} \), take the derivative of the top and bottom. Use the power rule for the part with \( n \). Then, put \( x=0 \) back into the new equation to find the answer.

🎯 Exam Tip: This limit is a classic result that can also be solved using the formula \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \). In this case, setting \( y = 1+x \), the limit becomes \( \lim_{y \to 1} \frac{y^n-1^n}{y-1} \).

Question 17. \( \lim _{x \rightarrow 0} \frac{1-x^{-1 / 3}}{1-x^{-2 / 3}} \)
Answer: Let's rewrite the expression using positive exponents first:
\( \lim _{x \rightarrow 0} \frac{1-\frac{1}{x^{1/3}}}{1-\frac{1}{x^{2/3}}} \)
Now, let's substitute \( x=0 \).
Numerator: \( 1-\frac{1}{0} \rightarrow 1-\infty \rightarrow -\infty \)
Denominator: \( 1-\frac{1}{0} \rightarrow 1-\infty \rightarrow -\infty \)
The limit is in \( \frac{\infty}{\infty} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \).
\( \frac{d}{dx}(1-x^{-1/3}) = 0 - (-\frac{1}{3})x^{-1/3-1} = \frac{1}{3}x^{-4/3} \)
\( \frac{d}{dx}(1-x^{-2/3}) = 0 - (-\frac{2}{3})x^{-2/3-1} = \frac{2}{3}x^{-5/3} \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{\frac{1}{3}x^{-4/3}}{\frac{2}{3}x^{-5/3}} \)
Simplify the expression:
\( = \lim _{x \rightarrow 0} \frac{1}{3} \times \frac{3}{2} \times \frac{x^{-4/3}}{x^{-5/3}} \)
\( = \lim _{x \rightarrow 0} \frac{1}{2} x^{-4/3 - (-5/3)} = \lim _{x \rightarrow 0} \frac{1}{2} x^{-4/3 + 5/3} \)
\( = \lim _{x \rightarrow 0} \frac{1}{2} x^{1/3} \)
Now, substitute \( x=0 \):
\( = \frac{1}{2} (0)^{1/3} = \frac{1}{2} \times 0 = 0 \)
Rewriting with positive exponents simplifies the differentiation and evaluation steps significantly.
In simple words: First, rewrite the terms to avoid negative powers. If you get \( \frac{\infty}{\infty} \), take the derivative of the top and bottom. Then, simplify the expression and put \( x=0 \) back in to find the answer.

🎯 Exam Tip: When dealing with negative or fractional exponents, it's often helpful to rewrite terms with positive exponents or radical form to avoid calculation errors during differentiation and simplification.

Question 18. \( \lim _{x \rightarrow 0} \frac{1-\cos x}{x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( 1-\cos 0 = 1-1 = 0 \)
Denominator: \( 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \).
\( \lim _{x \rightarrow 0} \frac{1-\cos x}{x} \)
\( = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x)} \)
\( = \lim _{x \rightarrow 0} \frac{0 - (-\sin x)}{1} \)
\( = \lim _{x \rightarrow 0} \sin x \)
Now, we substitute \( x=0 \) into the new expression:
\( = \sin 0 = 0 \)
This limit is a fundamental result in calculus, often used to derive the derivative of sine.
In simple words: If you get \( \frac{0}{0} \) from the limit, take the derivative of the top and bottom. Remember that the derivative of \( \cos x \) is \( -\sin x \). Then, put \( x=0 \) back into the new equation to find the answer.

🎯 Exam Tip: This is a very important standard limit. Knowing that \( \lim_{x \to 0} \frac{1-\cos x}{x} = 0 \) can save time, but understanding how to derive it using L'Hospital's rule is essential.

Question 19. \( \lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( 1-\cos(m \cdot 0) = 1-\cos 0 = 1-1 = 0 \)
Denominator: \( 1-\cos(n \cdot 0) = 1-\cos 0 = 1-1 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \). Remember the chain rule for \( \cos(ax) \).
\( \frac{d}{dx}(1-\cos mx) = 0 - (-\sin mx)(m) = m \sin mx \)
\( \frac{d}{dx}(1-\cos nx) = 0 - (-\sin nx)(n) = n \sin nx \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{m \sin mx}{n \sin nx} \)
Let's check the form again by substituting \( x=0 \).
Numerator: \( m \sin(m \cdot 0) = m \sin 0 = 0 \)
Denominator: \( n \sin(n \cdot 0) = n \sin 0 = 0 \)
It's still in \( \frac{0}{0} \) indeterminate form, so we apply L'Hospital's Rule again.
\( \frac{d}{dx}(m \sin mx) = m (\cos mx)(m) = m^2 \cos mx \)
\( \frac{d}{dx}(n \sin nx) = n (\cos nx)(n) = n^2 \cos nx \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{m^2 \cos mx}{n^2 \cos nx} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{m^2 \cos(m \cdot 0)}{n^2 \cos(n \cdot 0)} = \frac{m^2 \cos 0}{n^2 \cos 0} = \frac{m^2 \times 1}{n^2 \times 1} = \frac{m^2}{n^2} \)
This problem requires applying L'Hospital's rule twice.
In simple words: If you get \( \frac{0}{0} \) for a limit twice in a row after differentiating, just do L'Hospital's rule again. Remember the chain rule for \( \cos(mx) \). Keep differentiating until you can plug in the value and get a clear answer.

🎯 Exam Tip: For limits of the form \( \lim_{x \to 0} \frac{1-\cos(ax)}{1-\cos(bx)} \), a shortcut is \( \frac{a^2}{b^2} \). However, be prepared to show the double application of L'Hospital's Rule if asked for the full solution.

Question 20. \( \lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( \tan(2 \cdot 0)-0 = \tan 0 - 0 = 0-0 = 0 \)
Denominator: \( 3 \cdot 0-\sin 0 = 0-0 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \). Remember that \( \frac{d}{dx}(\tan(ax)) = a \sec^2(ax) \).
\( \frac{d}{dx}(\tan 2x-x) = 2 \sec^2 2x - 1 \)
\( \frac{d}{dx}(3x-\sin x) = 3 - \cos x \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{2 \sec^2 2x - 1}{3 - \cos x} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{2 \sec^2(2 \cdot 0) - 1}{3 - \cos 0} \)
\( = \frac{2 \sec^2 0 - 1}{3 - 1} \)
Since \( \sec 0 = \frac{1}{\cos 0} = \frac{1}{1} = 1 \):
\( = \frac{2(1)^2 - 1}{2} = \frac{2 - 1}{2} = \frac{1}{2} \)
This problem tests knowledge of derivatives of trigonometric functions and their values at 0.
In simple words: If you get \( \frac{0}{0} \) for a limit, take the derivative of the top and bottom. Remember how to differentiate \( \tan(2x) \) and \( \sin x \). Then, put \( x=0 \) back into the new equation to find the answer.

🎯 Exam Tip: Be careful with the chain rule for \( \tan(2x) \). The derivative is \( \sec^2(2x) \cdot \frac{d}{dx}(2x) = 2\sec^2(2x) \). Also, recall that \( \sec 0 = 1 \).

Question 21. \( \lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( e^{0^2}-\cos 0 = e^0-1 = 1-1 = 0 \)
Denominator: \( 0^2 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \).
\( \frac{d}{dx}(e^{x^2}-\cos x) = e^{x^2}(2x) - (-\sin x) = 2x e^{x^2} + \sin x \)
\( \frac{d}{dx}(x^2) = 2x \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{2x e^{x^2} + \sin x}{2x} \)
Let's check the form again by substituting \( x=0 \).
Numerator: \( 2(0) e^{0^2} + \sin 0 = 0+0 = 0 \)
Denominator: \( 2(0) = 0 \)
It's still in \( \frac{0}{0} \) indeterminate form, so we apply L'Hospital's Rule again.
\( \frac{d}{dx}(2x e^{x^2} + \sin x) = [2e^{x^2} + 2x \cdot e^{x^2}(2x)] + \cos x \) (using product rule for \( 2x e^{x^2} \))
\( = 2e^{x^2} + 4x^2 e^{x^2} + \cos x \)
\( \frac{d}{dx}(2x) = 2 \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{2e^{x^2} + 4x^2 e^{x^2} + \cos x}{2} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{2e^{0^2} + 4(0)^2 e^{0^2} + \cos 0}{2} \)
\( = \frac{2e^0 + 0 + 1}{2} = \frac{2(1) + 1}{2} = \frac{2+1}{2} = \frac{3}{2} \)
This problem requires multiple applications of L'Hospital's Rule and careful use of the product rule.
In simple words: If you get \( \frac{0}{0} \) after differentiating, do L'Hospital's rule again. Remember to use the product rule when differentiating terms like \( 2x e^{x^2} \). Keep going until you can plug in \( x=0 \) and get a number.

🎯 Exam Tip: Be very careful when applying the product rule for terms like \( 2x e^{x^2} \) during repeated differentiation. A common mistake is to forget the chain rule inside the product rule for \( e^{x^2} \).

Question 22. \( \lim _{x \rightarrow 0} \frac{x \cos x-\log (1+x)}{x^2} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( 0 \cdot \cos 0 - \log(1+0) = 0 \cdot 1 - \log 1 = 0 - 0 = 0 \)
Denominator: \( 0^2 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \).
\( \frac{d}{dx}(x \cos x - \log(1+x)) = (1 \cdot \cos x + x(-\sin x)) - \frac{1}{1+x} \) (using product rule for \( x \cos x \))
\( = \cos x - x \sin x - \frac{1}{1+x} \)
\( \frac{d}{dx}(x^2) = 2x \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{\cos x - x \sin x - \frac{1}{1+x}}{2x} \)
Let's check the form again by substituting \( x=0 \).
Numerator: \( \cos 0 - 0 \cdot \sin 0 - \frac{1}{1+0} = 1 - 0 - 1 = 0 \)
Denominator: \( 2(0) = 0 \)
It's still in \( \frac{0}{0} \) indeterminate form, so we apply L'Hospital's Rule again.
\( \frac{d}{dx}(\cos x - x \sin x - \frac{1}{1+x}) = -\sin x - (1 \cdot \sin x + x \cos x) - (-\frac{1}{(1+x)^2}) \)
\( = -\sin x - \sin x - x \cos x + \frac{1}{(1+x)^2} \)
\( = -2\sin x - x \cos x + \frac{1}{(1+x)^2} \)
\( \frac{d}{dx}(2x) = 2 \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{-2\sin x - x \cos x + \frac{1}{(1+x)^2}}{2} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{-2\sin 0 - 0 \cdot \cos 0 + \frac{1}{(1+0)^2}}{2} \)
\( = \frac{-2(0) - 0(1) + \frac{1}{1^2}}{2} = \frac{0 - 0 + 1}{2} = \frac{1}{2} \)
This problem requires careful application of the product rule and differentiation of logarithmic functions.
In simple words: If you get \( \frac{0}{0} \) twice after differentiating, apply L'Hospital's rule again. Remember the product rule for terms like \( x \cos x \) and the derivative of \( \log(1+x) \). Keep going until you can plug in \( x=0 \) to get a clear answer.

🎯 Exam Tip: When differentiating \( x \cos x \), apply the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Also, remember that \( \frac{d}{dx}(\ln(ax+b)) = \frac{a}{ax+b} \).

Question 23. \( \lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^2} \)
Answer: Let's check the form of the limit by substituting \( x=0 \).
Numerator: \( \cos(a \cdot 0) - \cos(b \cdot 0) = \cos 0 - \cos 0 = 1-1 = 0 \)
Denominator: \( 0^2 = 0 \)
The limit is in \( \frac{0}{0} \) indeterminate form.
We apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to \( x \).
\( \frac{d}{dx}(\cos ax - \cos bx) = -a \sin ax - (-b \sin bx) = -a \sin ax + b \sin bx \)
\( \frac{d}{dx}(x^2) = 2x \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{-a \sin ax + b \sin bx}{2x} \)
Let's check the form again by substituting \( x=0 \).
Numerator: \( -a \sin(a \cdot 0) + b \sin(b \cdot 0) = -a(0) + b(0) = 0 \)
Denominator: \( 2(0) = 0 \)
It's still in \( \frac{0}{0} \) indeterminate form, so we apply L'Hospital's Rule again.
\( \frac{d}{dx}(-a \sin ax + b \sin bx) = -a(a \cos ax) + b(b \cos bx) \)
\( = -a^2 \cos ax + b^2 \cos bx \)
\( \frac{d}{dx}(2x) = 2 \)
So, the limit becomes:
\( \lim _{x \rightarrow 0} \frac{-a^2 \cos ax + b^2 \cos bx}{2} \)
Now, we substitute \( x=0 \) into the new expression:
\( = \frac{-a^2 \cos(a \cdot 0) + b^2 \cos(b \cdot 0)}{2} \)
\( = \frac{-a^2 \cos 0 + b^2 \cos 0}{2} = \frac{-a^2(1) + b^2(1)}{2} \)
\( = \frac{b^2-a^2}{2} \)
This problem requires applying L'Hospital's rule twice and using the chain rule for trigonometric derivatives.
In simple words: If you get \( \frac{0}{0} \) twice after differentiating, apply L'Hospital's rule again. Remember the chain rule for \( \cos(ax) \) and \( \sin(ax) \). Keep differentiating until you can plug in \( x=0 \) to get a clear answer.

🎯 Exam Tip: For limits of the form \( \lim_{x \to 0} \frac{\cos(ax)-\cos(bx)}{x^2} \), a useful shortcut result is \( \frac{b^2-a^2}{2} \). This can be derived by applying L'Hospital's Rule twice.

Question 27. Find the limit of \( \log _{\sin 2 x} \sin x \) as \( x \rightarrow 0^{+} \).
Answer: First, we rewrite the logarithm using the change of base formula, making it a ratio of natural logarithms:
\[ \lim _{x \rightarrow 0^{+}} \frac{\log \sin x}{\log \sin 2x} \] This gives us the indeterminate form \( \frac{\infty}{\infty} \). So, we can apply L'Hospital's rule by differentiating the numerator and the denominator separately with respect to \( x \):
\[ \lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\sin x} \cos x}{\frac{1}{\sin 2x} \cdot 2 \cos 2x} \] Now, simplify the expression:
\[ \lim _{x \rightarrow 0^{+}} \frac{\sin 2x \cos x}{2 \sin x \cos 2x} \] We can split this into two parts for easier evaluation:
\[ \lim _{x \rightarrow 0^{+}} \frac{\sin 2x}{2 \sin x} \cdot \lim _{x \rightarrow 0^{+}} \frac{\cos x}{\cos 2x} \] The second limit evaluates to \( \frac{\cos 0}{\cos 0} = \frac{1}{1} = 1 \). For the first limit, we still have the indeterminate form \( \frac{0}{0} \). We apply L'Hospital's rule again:
\[ \frac{1}{2} \lim _{x \rightarrow 0^{+}} \frac{2 \cos 2x}{\cos x} \] Evaluate this limit:
\[ \frac{1}{2} \cdot \frac{2 \cos 0}{\cos 0} = \frac{1}{2} \cdot \frac{2 \cdot 1}{1} = 1 \] Thus, the final limit is \( 1 \). It's helpful to remember that logarithmic limits often involve rewriting or L'Hospital's rule.
In simple words: To solve this problem, we first changed the logarithm into a fraction using `log` (natural logarithm). Since this gave us a `infinity/infinity` form, we used L'Hospital's rule twice by differentiating the top and bottom parts. After simplifying and checking the limit again, we found the answer to be 1.

🎯 Exam Tip: When evaluating limits of logarithmic functions, converting them to a ratio of natural logarithms using the change of base formula is often the first crucial step to apply L'Hospital's rule.

Question 28. Find the limit of \( \left(\frac{1}{x^2}-\frac{\cot x}{x}\right) \) as \( x \rightarrow 0 \).
Answer: First, we combine the terms into a single fraction:
\[ \lim _{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{\cot x}{x}\right) = \lim _{x \rightarrow 0} \frac{x - x^2 \cot x}{x^3} \] This can be rewritten using \( \cot x = \frac{1}{\tan x} \):
\[ \lim _{x \rightarrow 0} \frac{x - \frac{x^2}{\tan x}}{x^3} = \lim _{x \rightarrow 0} \frac{x \tan x - x^2}{x^3 \tan x} \] We can separate this into two limits, knowing that \( \lim_{x \to 0} \frac{x}{\tan x} = 1 \):
\[ \lim _{x \rightarrow 0} \frac{\tan x - x}{x^3} \cdot \lim _{x \rightarrow 0} \frac{x}{\tan x} = \lim _{x \rightarrow 0} \frac{\tan x - x}{x^3} \cdot 1 \] Now, we focus on \( \lim _{x \rightarrow 0} \frac{\tan x - x}{x^3} \), which is in the indeterminate form \( \frac{0}{0} \). We apply L'Hospital's rule:
\[ \lim _{x \rightarrow 0} \frac{\sec^2 x - 1}{3x^2} \] This is still an indeterminate form \( \frac{0}{0} \) because \( \sec^2 0 - 1 = 1 - 1 = 0 \). We apply L'Hospital's rule again:
\[ \lim _{x \rightarrow 0} \frac{2 \sec x (\sec x \tan x)}{6x} = \lim _{x \rightarrow 0} \frac{2 \sec^2 x \tan x}{6x} \] We can split this limit and use the known limit \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):
\[ \lim _{x \rightarrow 0} \frac{2 \sec^2 x}{6} \cdot \lim _{x \rightarrow 0} \frac{\tan x}{x} \] Evaluating the limits:
\[ \frac{2 \sec^2 0}{6} \cdot 1 = \frac{2 \cdot 1^2}{6} \cdot 1 = \frac{2}{6} = \frac{1}{3} \] Thus, the final answer is \( \frac{1}{3} \). When dealing with limits that combine different functions, finding a common denominator is key.
In simple words: First, we put the two terms into one fraction. This gave us a `0/0` form, so we used L'Hospital's rule. We had to use it two times to solve the problem. After simplifying, we found the answer to be `1/3`.

🎯 Exam Tip: For limits involving differences like \( \frac{1}{x^2} - \frac{\cot x}{x} \), always combine them into a single fraction first to reveal a standard indeterminate form for L'Hospital's rule.