证明 \(\lim_{x\rightarrow0}\frac{(e^{sinx}+sinx)^{\frac{1}{sinx}}-(e^{tanx}+tanx)^{\frac{1}{tanx}}}{x^3}=\frac34 e^2\)
解答:
\[\begin{align} 记f(x)=(e^x+x)^{\frac{1}{x}}，则原式&=\lim_{x\rightarrow 0} \frac{f(\sin x)-f(\tan x)}{x^3} \\由拉格朗日中值定理，原式&=\lim_{\xi \rightarrow 0}\frac{f^\prime(\xi)(\sin x -\tan x)}{x^3} \\&=-\frac12\lim_{\xi \rightarrow 0}f^\prime(\xi) \\现计算\lim_{x \rightarrow 0}f^\prime(x), \lim_{x \rightarrow 0}f^\prime(x)&=(e^x+x)^{\frac{1}{x}}{[\frac{\ln (e^x+x)}{x}]}^\prime \\&=\lim_{x\rightarrow 0}(e^x+x)^{\frac{1}{x}} \lim_{x \rightarrow0}[-\frac{1}{x^2}\ln (x+e^x)+\frac{1+e^x}{x(x+e^x)}] \\&=e^2\lim_{x\rightarrow 0}\frac{\frac{e^x+1}{e^x+x}x-2x}{x^2}+e^2\lim_{x\rightarrow 0}\frac{2x-\ln(e^x+x)}{x^2} \\由洛必达法则，上式&=e^2\lim_{x\rightarrow0}\frac{\frac{e^x+1}{e^x+x}-2}{x}+e^2\lim_{x\rightarrow0}\frac{2-\frac{e^x+1}{e^x+x}}{2x} \\&=e^2\lim_{x\rightarrow0}\frac{\frac{e^x+1}{e^x+x}-2}{2x}=-\frac32e^2 \\ \Longrightarrow原式=(-\frac12)×(-\frac32&e^2)=\frac34e^2,原题即证 \end{align}\]