题目
1 计算下列极限1) lim _(xarrow 0)dfrac (sqrt {1+xsin x)-sqrt (cos x)}(xtan x);1) lim _(xarrow 0)dfrac (sqrt {1+xsin x)-sqrt (cos x)}(xtan x)
1 计算下列极限
;
题目解答
答案
极限
利用等价无穷小可得
∴
解析
步骤 1：计算极限 $\lim _{x\rightarrow 0}\dfrac {\sqrt {1+x\sin x}-\sqrt {\cos x}}{x\tan x}$
利用等价无穷小可得
$\tan x\sim x$
∴$\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-\sqrt {\cos x}}{x\tan x}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-\sqrt {\cos x}}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-\sqrt {1+\cos x-1}}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-1}{x^2}-\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+\cos x-1}-1}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}x\sin x}{x^2}-\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}(\cos x-1)}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}x^2}{x^2}+\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}\times \dfrac {1}{2}x^2}{x^2}$
$=\dfrac {1}{2}+\dfrac {1}{4}=\dfrac {3}{4}$
步骤 2：计算极限 $\lim _{x\rightarrow 0}$ $\dfrac {\sin (a+2x)-2\sin (a+x)+\sin a}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {(\sin (a+2x)-\sin (a+x))-(\sin (a+x)-\sin a)}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {2\cos (a+\dfrac {3x}{2})\sin \dfrac {x}{2}-2\cos (a+\dfrac {x}{2})\sin \dfrac {x}{2}}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {2\sin \dfrac {x}{2}(\cos (a+\dfrac {3x}{2})-\cos (a+\dfrac {x}{2}))}{x^2}$
$=-\lim _{x\rightarrow 0}$ $\dfrac {2\sin \dfrac {x}{2}(2\sin \dfrac {x}{2}\cdot \sin (a+x))}{x^2}$
$=-\lim _{x\rightarrow 0}$ $\dfrac {(2\sin \dfrac {x}{2})^2\sin (a+x)}{x^2}$
$=-\lim _{x\rightarrow 0}$ $\dfrac {x^2\sin (a+x)}{x^2}=-\sin a$
利用等价无穷小可得
$\tan x\sim x$
∴$\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-\sqrt {\cos x}}{x\tan x}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-\sqrt {\cos x}}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-\sqrt {1+\cos x-1}}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+x\sin x}-1}{x^2}-\lim _{x\rightarrow 0}$ $\dfrac {\sqrt {1+\cos x-1}-1}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}x\sin x}{x^2}-\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}(\cos x-1)}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}x^2}{x^2}+\lim _{x\rightarrow 0}$ $\dfrac {\dfrac {1}{2}\times \dfrac {1}{2}x^2}{x^2}$
$=\dfrac {1}{2}+\dfrac {1}{4}=\dfrac {3}{4}$
步骤 2：计算极限 $\lim _{x\rightarrow 0}$ $\dfrac {\sin (a+2x)-2\sin (a+x)+\sin a}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {(\sin (a+2x)-\sin (a+x))-(\sin (a+x)-\sin a)}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {2\cos (a+\dfrac {3x}{2})\sin \dfrac {x}{2}-2\cos (a+\dfrac {x}{2})\sin \dfrac {x}{2}}{x^2}$
$=\lim _{x\rightarrow 0}$ $\dfrac {2\sin \dfrac {x}{2}(\cos (a+\dfrac {3x}{2})-\cos (a+\dfrac {x}{2}))}{x^2}$
$=-\lim _{x\rightarrow 0}$ $\dfrac {2\sin \dfrac {x}{2}(2\sin \dfrac {x}{2}\cdot \sin (a+x))}{x^2}$
$=-\lim _{x\rightarrow 0}$ $\dfrac {(2\sin \dfrac {x}{2})^2\sin (a+x)}{x^2}$
$=-\lim _{x\rightarrow 0}$ $\dfrac {x^2\sin (a+x)}{x^2}=-\sin a$