📝 题目
例 6 求 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}\left\lbrack {\left( {{x}^{3} - {x}^{2} + \frac{x}{2}}\right) {\mathrm{e}}^{\frac{1}{x}} - \sqrt{{x}^{6} + 1}}\right\rbrack$ .
💡 答案与解析
解 ${\mathrm{e}}^{\frac{1}{x}} = 1 + \frac{1}{x} + \frac{1}{2{x}^{2}} + \frac{1}{6{x}^{3}} + o\left( \frac{1}{{x}^{3}}\right)$ ,
$$ \sqrt{{x}^{6} + 1} = {x}^{3}{\left( 1 + \frac{1}{{x}^{6}}\right) }^{\frac{1}{2}} = {x}^{3}\left\lbrack {1 + \frac{1}{2{x}^{6}} + o\left( \frac{1}{{x}^{6}}\right) }\right\rbrack , $$
$$ \left( {{x}^{3} - {x}^{2} + \frac{x}{2}}\right) {\mathrm{e}}^{\frac{1}{x}} - \sqrt{{x}^{6} + 1} = \frac{1}{12x} + \frac{1}{{12}{x}^{2}} - \frac{1}{2{x}^{3}} + \frac{1}{6} + o\left( 1\right) , $$
由此得 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}\left\lbrack {\left( {{x}^{3} - {x}^{2} + \frac{x}{2}}\right) {\mathrm{e}}^{\frac{1}{x}} - \sqrt{{x}^{6} + 1}}\right\rbrack = \frac{1}{6}$ .
评注 本题如果用洛必达法则计算, 则比较繁.