Question

Match List-I with List-II

\[ \begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A) \; \lim_{x \to 0} \frac{\ln(1+x)}{\sin x} & (I) \; 1 \\ (B) \; \lim_{x \to \infty} 2x \tan\left(\tfrac{1}{x}\right) & (II) \; 0 \\ (C) \; \lim_{x \to \infty} \frac{x^2}{e^x} & (III) \; 2 \\ (D) \; \lim_{x \to 1} x^{\tfrac{1}{x-1}} & (IV) \; e \\ \hline \end{array} \]
Choose the correct answer from the options given below:

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
• (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
• (A) - (IV), (B) - (II), (C) - (III), (D) - (I)

Hint:

For limits, quickly classify the indeterminate form: \(0/0, \infty/\infty, 0 \cdot \infty, \infty - \infty, 1^\infty, 0^0, \infty^0\).
For \(0/0\) or \(\infty/\infty\), use L'Hôpital's Rule or standard limits/series expansions.
For \(1^\infty, 0^0, \infty^0\), take the logarithm first to turn it into a \(0 \cdot \infty\) form, then rearrange for L'Hôpital's Rule.

Answer 1

The Correct Option is B

Step 1: Conceptual Overview:
We will evaluate four limits. Most of these limits present indeterminate forms that can be resolved using standard limit properties or L'Hôpital’s Rule.
Step 2: Detailed Analysis:
(A) \( \lim_{x \to 0} \dfrac{\ln(1+x)}{\sin x} \):
This limit results in the indeterminate form \(\tfrac{0}{0}\).
We can solve this by dividing the numerator and denominator by \(x\): \[\lim_{x\to 0} \frac{\ln(1+x)}{\sin x} = \lim_{x\to 0} \frac{\ln(1+x)/x}{\sin x / x} = \frac{1}{1} = 1\] Alternatively, applying L'Hôpital’s Rule: \[\lim_{x\to 0} \frac{1/(1+x)}{\cos x} = \frac{1}{1} = 1\] Therefore, (A) corresponds to (I).

(B) \( \lim_{x \to \infty} 2x \tan(1/x) \):
This limit is of the \(\infty \cdot 0\) form. Let \(u = 1/x\). As \(x \to \infty\), \(u \to 0\).
The limit becomes: \[\lim_{x\to\infty} 2x \tan(1/x) = \lim_{u\to 0} \frac{2\tan u}{u} = 2 \cdot 1 = 2\] Thus, (B) corresponds to (III).

(C) \( \lim_{x \to \infty} \dfrac{x^2}{e^x} \):
This limit presents the indeterminate form \(\tfrac{\infty}{\infty}\). Applying L'Hôpital’s Rule twice: \[\lim_{x\to\infty} \frac{2x}{e^x} \quad \text{Applying L'Hôpital's Rule again} \quad \to \quad \lim_{x\to\infty} \frac{2}{e^x} = 0\] Hence, (C) corresponds to (II).

(D) \( \lim_{x \to 1} x^{1/(x-1)} \):
This limit is of the \(1^\infty\) form. Let \(L = \lim_{x\to 1} x^{1/(x-1)}\). Taking the natural logarithm of both sides: \[\ln L = \lim_{x\to 1} \ln(x^{1/(x-1)}) = \lim_{x\to 1} \frac{\ln x}{x-1}\] This is a \(\tfrac{0}{0}\) form. Applying L'Hôpital’s Rule: \[\ln L = \lim_{x\to 1} \frac{1/x}{1} = 1 \quad \implies \quad L = e^1 = e\] Therefore, (D) corresponds to (IV).

Step 3: Conclusion:
The determined pairings are: - (A) → (I) - (B) → (III) - (C) → (II) - (D) → (IV) This matches Option (B).