2021年考研数二第17题解析：极限的计算、变上限积分、泰勒公式、洛必达运算

一、题目

难度评级：

二、解析

解法 1

由泰勒公式可知：

$$
\begin{aligned}
\int_{0}^{x}\mathrm{e}^{t^{2}}\mathrm{~d}t & = \int_{0}^{x}\left(1+t^{2}+o\left(t^{2}\right)\right)\mathrm{d}t=x+\frac{1}{3}x^{3}+o\left(x^{3}\right) \\ \\
\sin x & = x-\frac{1}{3!}x^{3}+o\left(x^{3}\right) \\ \\
\mathrm{e}^{x} & = 1 + x + \frac{1}{2!}x^{2} + o\left(x^{2}\right)
\end{aligned}
$$

所以：

$$
\begin{aligned}
& \ \lim_{x \to 0}\left(\frac{1+\int_{0}^{x}\mathrm{e}^{t^{2}}\mathrm{~d}t}{\mathrm{e}^{x}-1}-\frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\frac{\left(1+\int_{0}^{x}\mathrm{e}^{t^{2}}\mathrm{~d}t\right)\sin x-\left(\mathrm{e}^{x}-1\right)}{\left(\mathrm{e}^{x}-1\right)\sin x} \\ \\
= & \ \lim_{x \to 0}\frac{\left(1+x+\frac{1}{3}x^{3}+o\left(x^{3}\right)\right)\left(x-\frac{1}{3!}x^{3}+o\left(x^{3}\right)\right)-\left(x+\frac{1}{2!}x^{2}+o\left(x^{2}\right)\right)}{x^{2}+o\left(x^{2}\right)} \\ \\
= & \ \lim_{x \to 0}\frac{\left( 1+x \right) x – \left(x+\frac{1}{2}x^{2}\right)}{x^{2}} \\ \\
= & \ \lim_{x \to 0}\frac{\frac{1}{2}x^{2}}{x^{2}} \\ \\
= & \ \frac{1}{2}
\end{aligned}
$$

解法 2

由泰勒公式可知：

$$
\int_{0}^{x}\mathrm{e}^{t^{2}}\mathrm{~d}t = \int_{0}^{x}\left(1+t^{2}+o\left(t^{2}\right)\right)\mathrm{d}t=x+\frac{1}{3}x^{3}+o\left(x^{3}\right)
$$

于是：

$$
\begin{aligned}
& \ \lim_{x \to 0}\left(\frac{1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\left[\frac{1 + x + \frac{x^{3}}{3} + o\left(x^{3}\right)}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right] \\ \\
= & \ \lim_{x \to 0}\left(\frac{1 + x}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\frac{x}{\mathrm{e}^{x} – 1} + \lim_{x \to 0}\left(\frac{1}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ 1 + \lim_{x \to 0}\frac{\sin x – \mathrm{e}^{x} + 1}{\left(\mathrm{e}^{x} – 1\right)\sin x} \\ \\
= & \ 1 + \lim_{x \to 0}\frac{\sin x – \mathrm{e}^{x} + 1}{x^{2}} \\ \\
= & \ 1 + \lim_{x \to 0}\frac{\cos x – \mathrm{e}^{x}}{2x} \\ \\
= & \ 1 + \lim_{x \to 0}\frac{-\sin x – \mathrm{e}^{x}}{2} \\ \\
= & \ 1 + \frac{ \textcolor{lightgreen}{-0} – 1}{2} \\ \\
= & \ 1 + \frac{ \textcolor{lightgreen}{+0} – 1}{2} \\ \\
= & \ 1 + \frac{ \textcolor{lightgreen}{0} – 1}{2} \\ \\
= & \ 1 + \frac{ – 1}{2} \\ \\
= & \ 1 – \frac{1}{2} \\ \\
= & \ \frac{1}{2}
\end{aligned}
$$

解法 3

由等价无穷小公式，可知：

$$
\begin{aligned}
\mathrm{e}^{x} – 1 & \sim x \\
\sin x & \sim x \\
\sin x – x & \sim \frac{-1}{6} x^{3}
\end{aligned}
$$

所以：

$$
\begin{aligned}
& \ \lim_{x \to 0}\left(\frac{1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\frac{\left(1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t\right)\sin x – \mathrm{e}^{x} + 1}{\left(\mathrm{e}^{x} – 1\right)\sin x} \\ \\
= & \ \lim_{x \to 0}\frac{\left(1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t\right)\sin x – \mathrm{e}^{x} + 1}{x^{2}} \\ \\
= & \ \lim_{x \to 0}\left(\textcolor{pink}{ \frac{\sin x – x}{x^{2}} } + \frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t \cdot \sin x – \mathrm{e}^{x} + 1 + x}{x^{2}}\right) \\ \\
= & \ \textcolor{pink}{0} + \lim_{x \to 0}\frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t \cdot \sin x – \mathrm{e}^{x} + 1 + x}{x^{2}} \\ \\
= & \ \lim_{x \to 0}\frac{\sin x}{x} \cdot \frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{x} – \lim_{x \to 0}\frac{\mathrm{e}^{x} – 1 – x}{x^{2}} \\ \\
= & \ \lim_{x \to 0} \mathrm{e}^{x^{2}} – \lim_{x \to 0}\frac{\mathrm{e}^{x} – 1}{2x} \\ \\
= & \ 1 – \frac{1}{2} \\ \\
= & \ \frac{1}{2}
\end{aligned}
$$

解法 4

$$
\begin{aligned}
& \ \lim_{x \to 0}\left(\frac{1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\left(\frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{\mathrm{e}^{x} – 1} + \frac{1}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\left(\frac{\mathrm{e}^{x^{2}}}{\mathrm{e}^{x}} + \frac{1}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\left(1 + \frac{1}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\left(1 + \frac{\sin x – \mathrm{e}^{x} + 1}{\left(\mathrm{e}^{x} – 1\right)\sin x} \right) \\ \\
= & \ \lim_{x \to 0}\left(1 + \frac{1}{2} \cdot \frac{\sin x – \mathrm{e}^{x} + 1}{x^{2}} \right) \\ \\
= & \ \lim_{x \to 0}\left(1 + \frac{1}{2} \cdot \frac{\cos x – \mathrm{e}^{x}}{x} \right) \\ \\
= & \ \lim_{x \to 0}\left[1 + \frac{1}{2}\lim_{x \to 0}\left(-\sin x – \mathrm{e}^{x}\right) \right] \\ \\
= & \ \lim_{x \to 0}\left(1 – \frac{1}{2} \right) \\ \\
= & \ \frac{1}{2}
\end{aligned}
$$

解法 5

$$
\begin{aligned}
& \ \lim_{x \to 0}\left(\frac{1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\frac{\sin x + \sin x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t – \left(\mathrm{e}^{x} – 1\right)}{\left( \mathrm{e}^{x} – 1 \right) \sin x} \\ \\
= & \ \lim_{x \to 0}\frac{\sin x + \sin x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t – \left(\mathrm{e}^{x} – 1\right)}{x^{2}} \\ \\
= & \ \lim_{x \to 0}\frac{\cos x + \cos x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t + \sin x \cdot \mathrm{e}^{x^{2}} – \mathrm{e}^{x}}{2x} \\ \\
= & \ \lim_{x \to 0}\frac{\cos x + \cos x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t – \mathrm{e}^{x}}{2x} + \frac{\sin x \cdot \mathrm{e}^{x^{2}}}{2x} \\ \\
= & \ \lim_{x \to 0}\frac{\cos x + \cos x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t – \mathrm{e}^{x}}{2x} + \frac{1}{2} \\ \\
= & \ \lim_{x \to 0} \frac{1}{2} \left(-\sin x – \sin x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t + \cos x \cdot \mathrm{e}^{x^{2}} – \mathrm{e}^{x}\right) + \frac{1}{2} \\ \\
= & \ \frac{1}{2} \left( -0 – 0 + 1 – 1 \right) + \frac{1}{2} \\ \\
= & \ \frac{1}{2}
\end{aligned}
$$

解法 6

由等价无穷小公式，可知：

$$
\begin{aligned}
\mathrm{e}^{x} – 1 & \sim x \\
\sin x & \sim x \\
\cos x – 1 & \sim \frac{-1}{2} x^{2}
\end{aligned}
$$

所以：

$$
\begin{aligned}
& \ \lim_{x \to 0}\left(\frac{1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{\mathrm{e}^{x} – 1} – \frac{1}{\sin x}\right) \\ \\
= & \ \lim_{x \to 0}\frac{\sin x \left(1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t\right) – \mathrm{e}^{x} + 1}{\left(\mathrm{e}^{x} – 1\right)\sin x} \\ \\
= & \ \lim_{x \to 0}\frac{\sin x \left(1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t\right) – \mathrm{e}^{x} + 1}{x^{2}} \\ \\
= & \ \lim_{x \to 0}\frac{\cos x \left(1 + \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t\right) + \sin x \cdot \mathrm{e}^{x^{2}} – \mathrm{e}^{x}}{2x} \\ \\
= & \ \lim_{x \to 0}\frac{\cos x \textcolor{lightgreen}{- 1} + \cos x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t + \sin x \cdot \mathrm{e}^{x^{2}} – \mathrm{e}^{x} \textcolor{lightgreen}{+1}}{2x} \\ \\
= & \ \lim_{x \to 0}\frac{\cos x – 1}{2x} + \lim_{x \to 0}\frac{\cos x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{2x} + \lim_{x \to 0}\frac{\sin x \cdot \mathrm{e}^{x^{2}}}{2x} + \lim_{x \to 0}\frac{1 – \mathrm{e}^{x}}{2x} \\ \\
= & \ 0 + \frac{1}{2}\lim_{x \to 0}\cos x \cdot \lim_{x \to 0}\frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d}t}{x} + \frac{1}{2}\lim_{x \to 0}\frac{\sin x}{x} \cdot \lim_{x \to 0}\mathrm{e}^{x^{2}} + \frac{1}{2}\lim_{x \to 0}\frac{1 – \mathrm{e}^{x}}{x} \\ \\
= & \ 0 + \frac{1}{2} \times 1 \times 1 + \frac{1}{2} \times 1 \times 1 – \frac{1}{2} \\ \\
= & \ \frac{1}{2} + \frac{1}{2} – \frac{1}{2} \\ \\
= & \ \frac{1}{2}
\end{aligned}
$$

---