Key formulas: \[ \lim_{x\to 0}(1+ax)^{\frac{1}{x}}=e^{a},\qquad \lim_{x\to\infty}\left(1+\frac{c}{x}\right)^{kx}=e^{ck}. \] Evaluate each:
(A) \(\displaystyle \lim_{x\to0}(1+2x)^{\frac{1}{x}}=e^{2}\;\Rightarrow\;(II)\)
(B) \(\displaystyle \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e\;\Rightarrow\;(III)\)
(C) \(\displaystyle \lim_{x\to0}(1+5x)^{\frac{1}{x}}=e^{5}\;\Rightarrow\;(IV)\)
(D) \(\displaystyle \lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}=e^{3\cdot 2}=e^{6}\;\Rightarrow\;(I)\)

✅ Matching: \[ (A)\!\to\!(II),\quad (B)\!\to\!(III),\quad (C)\!\to\!(IV),\quad (D)\!\to\!(I). \]

We use the standard result: \[ \lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k \] Here, \(k = \tfrac{2}{3}\) and \(n = x\). \[ \therefore \;\; \lim_{x \to \infty} \left(1 + \frac{2}{3x}\right)^x = e^{\tfrac{2}{3}} \] Final Answer: \( e^{\tfrac{2}{3}} \)