📝 题目
例 2 设 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) = A$ ,求证: $\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\left\lbrack xf\left( x\right) \right\rbrack }{x} = A$ .
💡 答案与解析
证 不妨设 $x > 0$ ,注意到 ${xf}\left( x\right) - 1 < \left\lbrack {{xf}\left( x\right) }\right\rbrack \leq {xf}\left( x\right)$ ,有
$$ \frac{{xf}\left( x\right) - 1}{x} = f\left( x\right) - \frac{1}{x} < \frac{\left\lbrack xf\left( x\right) \right\rbrack }{x} \leq f\left( x\right) $$
$$ \overset{\text{ 由夹挤准则 }}{ = }\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\left\lbrack xf\left( x\right) \right\rbrack }{x} = A. $$