第227题 | 知识点：强化极限训练（十七）

题目

$$
\text{求极限 }\lim_{x\to+\infty}\sin\frac{1}{x}\cdot\int_x^{x^2}
(1+\frac{1}{2t})^t\sin\frac{1}{\sqrt{t}}dt
$$

解答

$$
\begin{aligned}
&
\lim_{x\to+\infty}\sin\frac{1}{x}\cdot\int_x^{x^2}
(1+\frac{1}{2t})^t\sin\frac{1}{\sqrt{t}}dt
\\
=&
\lim_{x\to+\infty}[\frac{\displaystyle\int_x^{x^2}
(1+\frac{1}{2t})^t\sin\frac{1}{\sqrt{t}}dt}{x}]
\\
\xlongequal{L’}&
\lim_{x\to+\infty}[2x(1+\frac{1}{2x^2})^{x^2}\cdot \sin\frac{1}{x} -
(1+\frac{1}{2x})^{x}\cdot \sin\frac{1}{\sqrt{x}}]
\\
=&
\lim_{x\to+\infty}[2 \cdot e^{x^2\ln(1+\frac{1}{2x^2})} -
e^{x\ln(1+\frac{1}{2x})}\cdot \frac{1}{\sqrt{x}}]
\\
=&
2\lim_{x\to+\infty}e^{x^2\ln(1+\frac{1}{2x^2})} - \lim_{x\to+\infty}
e^{x\ln(1+\frac{1}{2x})}\cdot \frac{1}{\sqrt{x}}
\\
=&
2e^{\frac{1}{2}} - 0
\\
=&
2e^{\frac{1}{2}}
\\
\end{aligned}
$$