微积分(A)随缘一题[2]

求：\(\lim_{x \to \infty}(\frac{1+\sqrt[x]{3}}{2})^x\)

设 \(\phi(x)=\frac{1+\sqrt[x]{3}}{2}-1\)

则 \(\lim_{x \to \infty} (1+\phi(x))^{\frac{1}{\phi(x)} \cdot x\phi(x)}=e^{\lim_{x \to \infty}x\frac{\sqrt[x]{3}-1}{2}}\)

考虑到 \(\lim_{x \to \infty} \frac{x}{2}(\sqrt[x]{3}-1)=\lim_{x \to \infty} \frac{x}{2}\frac{\sqrt[x]{3}-1}{\frac{1}{x}\ln 3} \cdot \frac{\ln 3}{x}=\frac{\ln 3}{2}\)

所以结果为：\(e^{\frac{\ln 3}{2}}=\sqrt{3}\)