Match List-I with List-II. $$\begin{array}{|c|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A)\ \lim_{x \to 0}(1+2x)^{\tfrac{1}{x}} & (I)\ e^{6} \\ \hline (B)\ \lim_{x \to \infty}(1+\tfrac{1}{x})^{x} & (II)\ e^{2} \\ \hline (C)\ \lim_{x \to 0}(1+5x)^{\tfrac{1}{x}} & (III)\ e \\ \hline (D)\ \lim_{x \to \infty}(1+\tfrac{3}{x})^{2x} & IV)\ e^{5} \\ \hline \end{array}$$
Step-by-step Solution:
We use the standard limit: \(\displaystyle\lim_{t\to 0}(1+at)^{1/t}=e^{a}\) and \(\displaystyle\lim_{n\to\infty}\Big(1+\frac{a}{n}\Big)^{bn}=e^{ab}.\)
(A) \(\displaystyle\lim_{x\to 0}(1+2x)^{1/x}\). Put \(a=2\) in the standard form \(\Rightarrow e^{2}\). So (A) → (II).
(B) \(\displaystyle\lim_{x\to\infty}\Big(1+\frac{1}{x}\Big)^{x}\). This is the defining limit of \(e\). So (B) → (III) (i.e. \(e\)).
(C) \(\displaystyle\lim_{x\to 0}(1+5x)^{1/x}\). Put \(a=5\) → \(e^{5}\). So (C) → (IV).
(D) \(\displaystyle\lim_{x\to\infty}\Big(1+\frac{3}{x}\Big)^{2x}\). Rewrite: \(\Big(1+\frac{3}{x}\Big)^{2x}=\Big[\Big(1+\frac{3}{x}\Big)^{x/3}\Big]^{6}\). As \(x\to\infty\), \(\Big(1+\frac{3}{x}\Big)^{x/3}\to e\), so the whole limit is \(e^{6}\). So (D) → (I).
Final matching:
(A) – (II), (B) – (III), (C) – (IV), (D) – (I)
Correct Option: (C)