📝 题目
例 7 求下列极限:
(1) $\mathop{\lim }\limits_{\substack{{x \rightarrow + \infty } \\ {y \rightarrow + \infty } }}{\left( \frac{xy}{{x}^{2} + {y}^{2}}\right) }^{x}$ ; (2) $\mathop{\lim }\limits_{\substack{{x \rightarrow \infty } \\ {y \rightarrow a} }}{\left( 1 + \frac{1}{x}\right) }^{\frac{{x}^{2}}{x + y}}$ .
💡 答案与解析
解 (1) 因 $0 \leq {\left( \frac{xy}{{x}^{2} + {y}^{2}}\right) }^{x} \leq {\left( \frac{1}{2}\right) }^{x}\left( {x > 0,y > 0}\right)$ ,所以
$$ \mathop{\lim }\limits_{\substack{{x \rightarrow + \infty } \\ {y \rightarrow + \infty } }}{\left( \frac{xy}{{x}^{2} + {y}^{2}}\right) }^{x} = 0. $$
(2) $\mathop{\lim }\limits_{\substack{{x \rightarrow \infty } \\ {y \rightarrow a} }}{\left( 1 + \frac{1}{x}\right) }^{\frac{{x}^{2}}{x + y}} = \mathop{\lim }\limits_{\substack{{x \rightarrow \infty } \\ {y \rightarrow a} }}{\left\lbrack {\left( 1 + \frac{1}{x}\right) }^{x}\right\rbrack }^{\frac{x}{x + y}} = {\mathrm{e}}^{1} = \mathrm{e}$ .