第230题 | 知识点：强化极限训练（二十）

题目

$$
\lim_{x\to0}\Big[\frac{1}{\ln(x+\sqrt{1+x^2})} - \frac{1}{\ln(1+x) + \int_0^xt(1+t)^{\frac{1}{t}}dt}\Big]
$$

解答

$$
\lim_{x\to0}\frac{\ln(x+\sqrt{1+x^2})}{x} = 1
\quad\Rightarrow\quad
\ln(x+\sqrt{1+x^2}) \sim x
$$

$$
\lim_{x\to0}\frac{\int_0^xt(1+t)^{\frac{1}{t}}dt}{x} = 0
\quad\Rightarrow\quad
\Big[ \ln(1+x) + \int_0^xt(1+t)^{\frac{1}{t}}dt\Big] \sim \ln(1+x) \sim x
$$

$$
\begin{aligned}
&
\lim_{x\to0}\Big[\frac{1}{\ln(x+\sqrt{1+x^2})} - \frac{1}{\ln(1+x) + \displaystyle\int_0^xt(1+t)^{\frac{1}{t}}dt}\Big]
\\
=&
\lim_{x\to0}\frac
{\ln(1 + x) + \displaystyle\int_0^xt(1 + t)^{\frac{1}{t}}dt - \ln(x+\sqrt{1+x^2})}
{x^2}
\\
=&
\lim_{x\to0}\Bigg[\frac{\ln(1+x) - \ln(x+\sqrt{1+x^2})}{x^2}\Bigg] +
\lim_{x\to0}\Bigg[\frac{\int_0^xt(1+t)^{\frac{1}{t}}dt}{x^2}\Bigg] \quad(极限的四则运算)
\\
=&
\lim_{x\to0}\Bigg[\frac{\dfrac{1}{\xi}(1 - \sqrt{1+x^2})}{x^2}\Bigg] +
\lim_{x\to0}\Bigg[\frac{x(1+x)^{\frac{1}{x}}}{2x}\Bigg] \quad\bigg(Lagrange中值定理\bigg)
\\
=&
\lim_{x\to0}\Bigg(\frac{1 \cdot (-1)}{2\sqrt{1+x^2}}\Bigg) +
\lim_{x\to0}\Bigg(\frac{e}{2}\Bigg) \quad(洛必达)
\\
=&
\frac{e - 1}{2}
\end{aligned}
$$