题目
三、计算题:本题共有8小题,其中 16-19 小题每小题7分, 20-23 小题每小题8分,共60-|||-分.计算题必须写出必要的计算过程,只写答案的不给分.-|||-16.计算极限 lim _(xarrow 0)dfrac (1)({x)^2}ln dfrac (sin x)(x)
题目解答
答案
解析
步骤 1：利用等价无穷小替换
$\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln \dfrac {\sin x}{x}=\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1+\dfrac {\sin x}{x}-1)$
步骤 2：利用泰勒展开式
$\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1+\dfrac {\sin x}{x}-1)=\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1+\dfrac {x-\frac {1}{6}{x}^{3}+o({x}^{3})}{x}-1)$
步骤 3：化简并求极限
$\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1-\frac {1}{6}{x}^{2}+o({x}^{2}))=\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}(-\frac {1}{6}{x}^{2}+o({x}^{2}))=-\frac {1}{6}$
$\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln \dfrac {\sin x}{x}=\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1+\dfrac {\sin x}{x}-1)$
步骤 2：利用泰勒展开式
$\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1+\dfrac {\sin x}{x}-1)=\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1+\dfrac {x-\frac {1}{6}{x}^{3}+o({x}^{3})}{x}-1)$
步骤 3：化简并求极限
$\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}\ln (1-\frac {1}{6}{x}^{2}+o({x}^{2}))=\lim _{x\rightarrow 0}\dfrac {1}{{x}^{2}}(-\frac {1}{6}{x}^{2}+o({x}^{2}))=-\frac {1}{6}$