题目

$$
\text{求极限 }\lim_{x\to0}\Bigg(\frac{1 + \displaystyle\int_0^xe^{t^2}dt}{e^x-1} - \frac{1}{\sin x}\Bigg)
$$

解答

$$
\begin{aligned}
\lim_{x\to0}\Bigg(\frac{1 + \displaystyle\int_0^xe^{t^2}dt}{e^x-1} - \frac{1}{\sin x}\Bigg)
&=
\lim_{x\to0}\Bigg(\frac{\displaystyle\int_0^xe^{t^2}dt}{e^x-1} - \frac{\sin x - e^x + 1}{(e^x-1)\sin x}\Bigg)
\\
&=
\lim_{x\to0} \frac{\displaystyle\int_0^xe^{t^2}dt}{e^x-1} - \lim_{x\to0}\frac{\sin x - e^x + 1}{(e^x-1)\sin x}
\\
&=
\lim_{x\to0} \frac{\displaystyle\int_0^xe^{t^2}dt}{x} - \lim_{x\to0}\frac{x - x - \dfrac{1}{2}x^2}{x^2}
\\
&=
1 - \dfrac{1}{2}
\\
&= \frac{1}{2}
\end{aligned}
$$