变限积分的极限怎么算？放缩法试一试哦！

一、题目

$$
\lim \limits_{x \rightarrow+\infty} \int_{x_{0}}^{x} \ln \left(1+\frac{1}{\sqrt{t}}\right) \mathrm{d} t = ?
$$

其中 $x_{0}>0$ 且 $x>x_{0}$.

难度评级：

二、解析

由题可知：

$$
\lim \limits_{x \rightarrow+\infty} \int_{x_{0}}^{x} \ln \left(1+\frac{1}{\sqrt{t}}\right) \mathrm{d} t \Rightarrow
$$

$$
x \rightarrow+\infty \Rightarrow \ln \left(1+\frac{1}{\sqrt{t}}\right) \sim \frac{1}{\sqrt{t}} \Rightarrow
$$

$$
x_{0}>0, x>x_{0} \Rightarrow \frac{1}{\sqrt{t}}>0 \Rightarrow
$$

$$
\ln \left(1+\frac{1}{\sqrt{t}}\right)>\frac{1}{2} \cdot \frac{1}{\sqrt{t}} \Rightarrow
$$

$$
\int_{x_{0}}^{x} \ln \left(1+\frac{1}{\sqrt{t}}\right) \mathrm{d} t>\frac{1}{2}\int_{x_{0}}^{x} \frac{1}{\sqrt{t}} \mathrm{d} t.
$$

又：

$$
\frac{1}{2} \int_{x_{0}}^{x} \frac{1}{\sqrt{t}} \mathrm{d} t=\left.\frac{1}{2} \cdot 2 t^{\frac{1}{2}}\right|_{x_{0}} ^{x} \Rightarrow \sqrt{x}-\sqrt{x_{0}} \Rightarrow
$$

$$
x>x_{0} \Rightarrow \lim \limits_{x \rightarrow+\infty}\left(\sqrt{x}-\sqrt{x_{0}}\right)=+\infty.
$$

于是：

$$
\int_{x_{0}}^{x} \ln \left(1+\frac{1}{\sqrt{t}}\right) \mathrm{d} t=+\infty
$$

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