📝 题目
2. 5.1 设 $f\left( x\right)$ 在 $\left( {a, + \infty }\right)$ 上可导,且 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}\left\lbrack {f\left( x\right) + x{f}^{\prime }\left( x\right) }\right\rbrack = l$ . 求证:
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) = l $$
💡 答案与解析
2.5.1 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) \overset{\text{ 写成 }}{ = }\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\mathrm{e}}^{x}f\left( x\right) }{{\mathrm{e}}^{x}}\overset{\text{ 洛必达法则 }}{ = }\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\left( {\mathrm{e}}^{x}f\left( x\right) \right) }^{\prime }}{{\left( {\mathrm{e}}^{x}\right) }^{\prime }}$ .