问题及解答

(1)

\[
\lim_{x\rightarrow 0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{\sin x^3}
\]

[Ans] $\frac{1}{4}$

(2) 

\[
\lim_{x\rightarrow 0}\biggl(\frac{\sin x}{x}\biggr)^{\frac{1}{1-\cos x}}
\]

\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\
&=\lim_{x\rightarrow 0}\frac{\sin x(\frac{1}{\cos x}-1)}{x^3\cdot 2}\\
&=\lim_{x\rightarrow 0}\frac{\frac{1-\cos x}{\cos x}}{2x^2}\\
&=\lim_{x\rightarrow 0}\frac{1-\cos x}{2x^2}\\
&=\lim_{x\rightarrow 0}\frac{\frac{1}{2}x^2}{2x^2}\\
&=\frac{1}{4}.
\end{split}
\]

(2)

\[
\text{原式}=\lim_{x\rightarrow 0}e^{\frac{1}{1-\cos x}\ln\frac{\sin x}{x}}=e^{\lim\limits_{x\rightarrow 0}\frac{1}{1-\cos x}\ln\frac{\sin x}{x}}
\]

而

\[
\begin{split}
\lim_{x\rightarrow 0}\frac{1}{1-\cos x}\ln\frac{\sin x}{x}&=\lim_{x\rightarrow 0}\frac{\ln\frac{\sin x}{x}}{1-\cos x}\\
&=\lim_{x\rightarrow 0}\frac{\ln\frac{\sin x}{x}}{\frac{1}{2}x^2}\\
&=\lim_{x\rightarrow 0}\frac{\frac{x}{\sin x}\cdot\frac{x\cos x-\sin x}{x^2}}{x}\\
&=\lim_{x\rightarrow 0}\frac{x\cos x-\sin x}{x^3}\\
&=\lim_{x\rightarrow 0}\frac{\cos x+x(-1)\sin x-\cos x}{3x^2}\\
&=\lim_{x\rightarrow 0}\frac{-x\sin x}{3x^2}\\
&=-\frac{1}{3}
\end{split}
\]

或者使用 $\sin x$ 的 Taylor 展式, $\sin x=x-\frac{1}{3!}x^3+o(x^4)$. 于是

\[
\ln\frac{\sin x}{x}=\ln\frac{x-\frac{1}{3!}x^3+o(x^4)}{x}=\ln(1-\frac{1}{6}x^2+o(x^3))
\]

\[
\lim_{x\rightarrow 0}\frac{\ln\frac{\sin x}{x}}{\frac{1}{2}x^2}=\lim_{x\rightarrow 0}\frac{\ln(1-\frac{1}{6}x^2+o(x^3))}{\frac{1}{2}x^2}=\lim_{x\rightarrow 0}\frac{-\frac{1}{6}x^2+o(x^3)}{\frac{1}{2}x^2}=-\frac{1}{3}
\]

因此, 原极限的值为 $e^{-\frac{1}{3}}$.