问题及解答

求极限

1.    $\lim\limits_{x\rightarrow 0}(1-\tan x)^{\cot 2x}$.

2.   $\lim\limits_{x\rightarrow 0}(\cos x)^{\frac{1}{x^2}}$

3.   $\lim\limits_{x\rightarrow 0}(1-\tan x)^{\cot x}$.

$\tan x\rightarrow 0$ (当 $x\rightarrow 0$), 又注意到

\[
\cot 2x=\frac{1}{\tan 2x}=\frac{1-\tan^2 x}{2\tan x}.
\]

因此,

\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow 0}\biggl[\Bigl(1+(-\tan x)\Bigr)^{\frac{1}{-\tan x}}\biggr]^{\frac{\tan^2 x-1}{2}}\\
&=e^{-\frac{1}{2}}.
\end{split}
\]

2. 

\[
\begin{split}
\lim_{x\rightarrow 0}(\cos x)^{\frac{1}{x^2}}&=\lim_{x\rightarrow 0}\Bigl(1+(\cos x-1)\Bigr)^{\frac{1}{x^2}}\\
&=\lim_{x\rightarrow 0}\biggl[\Bigl(1+(\cos x-1)\Bigr)^{\frac{1}{\cos x-1}}\biggr]^{\frac{\cos x-1}{x^2}}\\
&=\lim_{x\rightarrow 0}\biggl[\Bigl(1+(\cos x-1)\Bigr)^{\frac{1}{\cos x-1}}\biggr]^{\frac{-\frac{1}{2}x^2}{x^2}}\\
&=\lim_{x\rightarrow 0}\biggl[\Bigl(1+(\cos x-1)\Bigr)^{\frac{1}{\cos x-1}}\biggr]^{\frac{-1}{2}}\\
&=e^{\frac{-1}{2}}
\end{split}
\]

或者

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

\[
\lim_{x\rightarrow 0}(1-\tan x)^{\cot 2x}
\]

也可以这样算. 注意到 $x\rightarrow 0$ 时, $\tan x\rightarrow 0$, 从而 $1-\tan x > 0$, 于是

\[
(1-\tan x)^{\cot 2x}=e^{\cos 2x\cdot\ln(1-\tan x)}
\]

\[
\begin{split}
\lim_{x\rightarrow 0}(1-\tan x)^{\cot 2x}&=\lim_{x\rightarrow 0}e^{\cos 2x\cdot\ln(1-\tan x)}\\
&=\exp(\lim_{x\rightarrow 0}\frac{\ln(1+(-\tan x))}{\tan 2x})\\
&=\exp(\lim_{x\rightarrow 0}\frac{-\tan x}{2x})\\
&=\exp(\lim_{x\rightarrow 0}\frac{-x}{2x})\\
&=e^{-\frac{1}{2}}.
\end{split}
\]

(3)

\[
\lim_{x\rightarrow 0}(1-\tan x)^{\cot x}=\lim_{x\rightarrow 0}\Bigl[\bigl(1+(-\tan x)\bigr)^{\frac{1}{-\tan x}}\Bigr]^{-1}=e^{-1}.
\]

这里用到了重要极限 $\lim\limits_{u\rightarrow 0}(1+u)^{\frac{1}{u}}=e$.