题目
5、 lim _(xarrow 0)dfrac (ln (1+x)-sin x)(2)
题目解答
答案
本题考查极限的计算，通过等价无穷小可得答案。
$\lim _{x\rightarrow 0}\dfrac {\ln (1+x)-\sin x}{\sqrt {1+{x}^{2}}-\cos x}$
$=\lim _{x\rightarrow 0}\dfrac {\dfrac {1}{1+x}-\cos x}{\dfrac {x}{\sqrt {1+{x}^{2}}+\cos x}}$
$=\lim _{x\rightarrow 0}\dfrac {1-(1+x)\cos x}{{x}^{2}-\cos x(\sqrt {1+{x}^{2}}+\cos x)}$
$=\lim _{x\rightarrow 0}\dfrac {1-\cos x+o{x}^{2}-{x}^{2}}{2x\sin x+o({x}^{2})-2\cos x\sin x-x\cos x}$
$=\lim _{x\rightarrow 0}\dfrac {1-\cos x}{2\sin x}$
$=\lim _{x\rightarrow 0}\dfrac {2\sin ^{2}\dfrac {x}{2}}{2\sin x}$
$=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}\dfrac {x}{2}}{x\sin x}$
$=\lim _{x\rightarrow 0}\dfrac {x^{2}}{x^{3}}$
$=\dfrac {1}{3}$
$\lim _{x\rightarrow 0}\dfrac {\ln (1+x)-\sin x}{\sqrt {1+{x}^{2}}-\cos x}$
$=\lim _{x\rightarrow 0}\dfrac {\dfrac {1}{1+x}-\cos x}{\dfrac {x}{\sqrt {1+{x}^{2}}+\cos x}}$
$=\lim _{x\rightarrow 0}\dfrac {1-(1+x)\cos x}{{x}^{2}-\cos x(\sqrt {1+{x}^{2}}+\cos x)}$
$=\lim _{x\rightarrow 0}\dfrac {1-\cos x+o{x}^{2}-{x}^{2}}{2x\sin x+o({x}^{2})-2\cos x\sin x-x\cos x}$
$=\lim _{x\rightarrow 0}\dfrac {1-\cos x}{2\sin x}$
$=\lim _{x\rightarrow 0}\dfrac {2\sin ^{2}\dfrac {x}{2}}{2\sin x}$
$=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}\dfrac {x}{2}}{x\sin x}$
$=\lim _{x\rightarrow 0}\dfrac {x^{2}}{x^{3}}$
$=\dfrac {1}{3}$