求极限 $\lim _{n \rightarrow \infty}\left(\int_1^2 \sqrt[n]{1+x} \mathrm{~d} x\right)^n$.
【答案】 $\int_1^2 \sqrt[n]{1+x} \mathrm{~d} x=\frac{n}{n+1}\left(3^{1+\frac{1}{n}}-2^{1+\frac{1}{n}}\right)$, 从而
$$
\begin{gathered}
\lim _{n \rightarrow \infty}\left(\int_1^2 \sqrt[n]{1+x} \mathrm{~d} x\right)^n=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{-n}\left(3^{1+\frac{1}{n}}-2^{1+\frac{1}{n}}\right)^n=\mathrm{e}^{-1} \lim _{n \rightarrow \infty}\left(3^{1+\frac{1}{n}}-2^{1+\frac{1}{n}}\right)^n . \\
\lim _{x \rightarrow 0^{+}}\left(3 \cdot 3^x-2 \cdot 2^x\right)^{\frac{1}{x}}=\mathrm{e}^{\lim _{x \rightarrow 0^{+}} \dfrac{\ln \left(3 \cdot 3^x-2 \cdot 2^x\right)}{x}}=\mathrm{e}^{3 \ln 3-2 \ln 2}=\frac{27}{4}, \text { 故 }_{n \rightarrow \infty} \lim _{n \rightarrow \infty}\left(3^{1+\frac{1}{n}}-2^{1+\frac{1}{n}}\right)^n=\frac{27}{4} .
\end{gathered}
$$
因此 $\lim _{n \rightarrow \infty}\left(\int_1^2 \sqrt[n]{1+x} \mathrm{~d} x\right)^n=\frac{27}{4} \mathrm{e}^{-1}$.