一、题目
设函数 $f\left( x \right)$ 在 $x = 0$ 处连续，且 $\lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)}=-3$，证明：$f\left( x \right)$ 在 $x = 0$ 处可导，并求 $f^{\prime}\left( 0 \right)$.
难度评级：
二、解析
本题的求解需要使用泰勒公式和常用的等价无穷小公式.
方法 1
由 $\lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)}=-3$ 可得：
$$
\begin{align}
-3 & = \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)} \notag \\ \notag \\
& = \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right) \left( 1-x \right)} \notag \\ \notag \\
& = \lim_{{x \to 0}} \frac{x f\left( x \right) – \textcolor{orange}{ \mathrm{e}^{2 \sin x}} + 1}{\ln \left( 1-x^{2} \right)} \notag \\ \notag \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\mathrm{e}^{\square} = 1 + \square + \frac{1}{2} \square^{2} + o \left( \square^{2} \right)} \notag \\ \notag \\
& = \lim_{{x \to 0}} \frac{x f\left( x \right)-\left[ \textcolor{orange}{ 1+2 \sin x+\frac{1}{2}\left( 2 \sin x \right)^{2}+o\left( x^{2} \right) } \right]+1}{-x^{2}} \notag \\ \notag \\
& = \lim_{{x \to 0}} \frac{x f\left( x \right)-2 \sin x-2 \sin^{2} x+o\left( x^{2} \right)}{-x^{2}} \notag \\ \notag \\
& = \lim_{{x \to 0}} \frac{x f\left( x \right)-2 x-2 x^{2}+o\left( x^{2} \right)}{-x^{2}} \notag \\ \notag \\
& = 2-\lim_{{x \to 0}} \frac{f\left( x \right)-2}{x} \notag \\ \notag \\
& \textcolor{lightgreen}{ \leadsto } 2-\lim_{{x \to 0}} \frac{f\left( x \right)-2}{x} = -3 \notag \\ \notag \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{lightgreen}{ \lim_{{x \to 0}} \frac{f\left( x \right)-2}{x} = 5 } \tag{1}
\end{align}
$$
又因为函数 $f\left( x \right)$ 在 $x = 0$ 处连续，所以：
$$
f\left( 0 \right)=\lim_{{x \to 0}} f\left( x \right) = 2 \tag{2}
$$
综合上面的 $(1)$ 式和 $(2)$ 式，可得：
$$
f^{\prime}\left( 0 \right) = \lim_{{x \to 0}} \frac{f\left( x \right)-f\left( 0 \right)}{x} = \lim_{{x \to 0}} \frac{f\left( x \right)-2}{x} = 5
$$
方法 2
首先，对题目所给得式子 $\lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)}=-3$ 进行变形，得：
$$
\begin{aligned}
& \ \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)} \\ \\
= & \ \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1-x^{2} \right)} \\ \\
= & \ \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{-x^{2}} \\ \\
= & \ -3
\end{aligned}
$$
接着：
$$
\begin{aligned}
& \ \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{-x^{2}} = -3 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{{x \to 0}} \left[ \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{x^{2}} = \frac{3}{1} \right] \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{{x \to 0}} \left[ x f\left( x \right)-\mathrm{e}^{2 \sin x}+1 = 3 x^{2} \right] \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{{x \to 0}} \left[ x f\left( x \right) = \mathrm{e}^{2 \sin x}-1 + 3 x^{2} \right] \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{{x \to 0}} \left[ f\left( x \right) = \frac{\mathrm{e}^{2 \sin x}-1}{x} + 3 x + o \left( x \right) \right] \\ \\
\end{aligned}
$$
因此，由极限和无穷小的关系，可知：
$$
f\left( x \right)=\frac{\mathrm{e}^{2 \sin x}-1}{x}-\left( -3+\alpha \right)x
$$
其中 $\lim_{{x \to 0}} \alpha =0$, 所以 $\alpha x \sim o \left( x \right)$.
又由于 $f\left( x \right)$ 连续，所以：
$$
\begin{aligned}
f\left( 0 \right) & = \lim_{{x \to 0}} f\left( x \right) \\ \\
& = \lim_{{x \to 0}} \left[ \frac{\mathrm{e}^{2 \sin x}-1}{x}-\left( -3+\alpha \right)x \right] \\ \\
& = \lim_{{x \to 0}} \frac{\mathrm{e}^{2 \sin x}-1}{x} \\ \\
& = \lim_{{x \to 0}} \frac{2 \sin x}{x} \\ \\
& = 2
\end{aligned}
$$
于是：
$
\begin{aligned}
f^{\prime}\left( 0 \right) & = \lim_{{x \to 0}} \frac{f\left( x \right)-f\left( 0 \right)}{x} \\ \\
& = \lim_{{x \to 0}} \frac{\frac{\mathrm{e}^{2 \sin x}-1}{x}-\left( -3+\alpha \right)x-2}{x} \\ \\
& = \lim_{{x \to 0}} \frac{\mathrm{e}^{2 \sin x}-1-\left( -3+\alpha \right)x^{2}-2 x}{x^{2}} \\ \\
& = \lim_{{x \to 0}} \frac{\mathrm{e}^{2 \sin x}-1-2 x}{x^{2}}+3 \\ \\
& = \lim_{{x \to 0}} \frac{1+2 \sin x+\frac{1}{2}\left( 2 \sin x \right)^{2}+o\left( x^{2} \right)-1-2 x}{x^{2}}+3 \\ \\
& = \lim_{{x \to 0}} \frac{2 \sin x-2 x}{x^{2}}+\lim_{{x \to 0}} \frac{\frac{1}{2}\left( 2 \sin x \right)^{2}+o\left( x^{2} \right)}{x^{2}}+3 \\ \\
& = 0 + 2 + 3 = 5
\end{aligned}
$
方法 3
由泰勒公式，可知：
$$
\begin{aligned}
& \ \ln\left(1+x\right)+\ln\left(1-x\right) \\ \\
= & \ x-\frac{1}{2}x^{2}+o\left(x^{2}\right)-x-\frac{1}{2}x^{2}+o\left(x^{2}\right) \\ \\
= & \ -x^{2}+o\left(x^{2}\right) \\ \\ \\
& \ \mathrm{e}^{2 \sin x} \\ \\
= & \ 1+2 \sin x+\frac{1}{2}\left(2 \sin x\right)^{2}+o\left(x^{2}\right) \\ \\
= & \ 1+2 \sin x+2 \sin^{2} x+o\left(x^{2}\right)
\end{aligned}
$$
于是，由题可得：
$$
\begin{aligned}
-3 & = \lim_{x \to 0}\frac{x f\left(x\right)-\mathrm{e}^{2 \sin x}+1}{-x^{2}} \\ \\
& = \lim_{x \to 0}\frac{x f\left(x\right)-\left[1+2 \sin x+2 \sin^{2} x+o\left(x^{2}\right)\right]+1}{-x^{2}} \\ \\
& = \lim_{x \to 0}\frac{x f\left(x\right)-2 \sin x}{-x^{2}}+\lim_{x \to 0}\frac{-2 \sin^{2} x}{-x^{2}} \\ \\
& = \lim_{x \to 0}\frac{x f\left(x\right)-2 \sin x}{-x^{2}} – 2
\end{aligned}
$$
进而可得：
$$
\begin{aligned}
& \ \lim_{x \to 0}\frac{x f\left(x\right)-2 \sin x}{-x^{2}}=-5 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{x \to 0}\frac{x f\left(x\right)-2\left(x-\frac{1}{6}x^{3}+o\left(x^{3}\right)\right)}{x^{2}}=5 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{x \to 0}\frac{x f\left(x\right)-2x}{x^{2}}=5 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{x \to 0}\frac{f\left(x\right)-2}{x}=5 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{x \to 0}\left[f\left(x\right)-2\right]=0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim_{x \to 0} f\left(x\right)=2=f\left(0\right)
\end{aligned}
$$
综上可得：
$$
f^{\prime}\left(0\right)=\lim_{x \to 0}\frac{f\left(x\right)-2}{x}=5
$$
方法 4
由题可得：
$$
\begin{align}
-3 & = \lim_{x \to 0}\frac{x f\left(x\right)-\mathrm{e}^{2 \sin x}+1}{\ln\left(1+x\right)+\ln\left(1-x\right)} \notag \\ \notag \\
& = \lim_{x \to 0}\frac{x f\left(x\right)-\mathrm{e}^{2 \sin x}+1}{\ln\left(1-x^{2}\right)} \notag \\ \notag \\
& = \lim_{x \to 0}\frac{x f\left(x\right)-\mathrm{e}^{2 \sin x}+1}{-x^{2}} \tag{3}
\end{align}
$$
又因为：
$$
\lim_{x \to \infty} \left[ – x (0) + 2x \right] = 0 \tag{4}
$$
于是，结合题目给的式子 $-3 = \lim_{{x \to 0}} \frac{x f\left( x \right)-\mathrm{e}^{2 \sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)}$, 以及上面的 $(3)$ 式和 $(4)$ 式，可知：
$$
\begin{aligned}
-3 & = \lim_{x \to 0}\frac{x f\left(x\right) – x f\left(0\right) + 2x-\mathrm{e}^{2 \sin x}+1}{-x^{2}} \\ \\
& = \lim_{x \to 0}\frac{x \left[f\left(x\right)-f\left(0\right)\right]+2x-\mathrm{e}^{2 \sin x}+1}{-x^{2}} \\ \\
& = -\lim_{x \to 0}\frac{f\left(x\right)-f\left(0\right)}{x}+\lim_{x \to 0}\frac{2x-\mathrm{e}^{2 \sin x}+1}{x^{2}} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\text{洛必达运算}} \\ \\
& = -\lim_{x \to 0}\frac{f\left(x\right)-f\left(0\right)}{x}+\lim_{x \to 0}\frac{2 – \mathrm{e}^{2 \sin x} \times 2 \cos x }{-2x} \\ \\
& =-\lim_{x \to 0}\frac{f\left(x\right)-f\left(0\right)}{x}+\lim_{x \to 0}\frac{1-\cos x + \left(1-\mathrm{e}^{2 \sin x}\right) \times \cos x}{-x} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\lim_{x \to 0}\frac{1-\mathrm{e}^{2 \sin x}}{x}=-2} \\ \\
& = – \lim_{x \to 0}\frac{f\left(x\right)-f\left(0\right)}{x}+2 = -3
\end{aligned}
$$
综上可得：
$$
\lim_{x \to 0}\frac{f\left(x\right)-f\left(0\right)}{x}=5
$$
即 $f\left(x\right)$ 在 $x=0$ 处可导且 $f^{\prime}\left(0\right)=5$.
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