📝 题目
例 4 求极限 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}{\left( 1 + \frac{1}{x}\right) }^{{x}^{2}}{\mathrm{e}}^{-x}$ .
💡 答案与解析
解法 1 令 $f\left( x\right) = {\left( 1 + \frac{1}{x}\right) }^{{x}^{2}}{\mathrm{e}}^{-x}$ ,则有
$$ \ln f\left( x\right) = {x}^{2}\ln \left( {1 + \frac{1}{x}}\right) - x\overset{t = \frac{1}{x}}{ = }\frac{\ln \left( {1 + t}\right) - t}{{t}^{2}}. $$
注意到当 $\displaystyle{x \rightarrow + \infty}$ 时, $t \rightarrow 0 + 0$ . 故有
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\ln f\left( x\right) = \mathop{\lim }\limits_{{t \rightarrow 0 + 0}}\frac{\ln \left( {1 + t}\right) - t}{{t}^{2}}\text{ ，洛必达法则 }\mathop{\lim }\limits_{{t \rightarrow 0 + 0}}\frac{\frac{1}{1 + t} - 1}{2t} = - \frac{1}{2}. $$
因此 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\mathrm{e}}^{\ln f\left( x\right) } = {\mathrm{e}}^{-\frac{1}{2}}$ .
解法 2 注意到当 $\displaystyle{x \rightarrow + \infty}$ 时, $\frac{1}{x}$ 是无穷小量.
$$ \ln f\left( x\right) = {x}^{2}\ln \left( {1 + \frac{1}{x}}\right) - x\overset{\text{ 泰勒公式 }}{ = }{x}^{2}\left\lbrack {\frac{1}{x} - \frac{1}{2{x}^{2}} + o\left( \frac{1}{{x}^{2}}\right) }\right\rbrack - x $$
$$ = - \frac{1}{2} + o\left( 1\right) . $$
由此即得 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\mathrm{e}}^{\ln f\left( x\right) } = {\mathrm{e}}^{-\frac{1}{2}}$ .