题目
8. lim _(xarrow 0)dfrac (e-{e)^cos x}(sqrt [3]{{x)^2+1}-1}
题目解答
答案
本题考查了极限的求。
：$\lim _{x\rightarrow 0}\dfrac {e-{e}^{\cos x}}{\sqrt [3]{{x}^{2}+1}-1}$
$=\lim _{x\rightarrow 0}\dfrac {e({e}^{\cos x}-1)}{{e}^{\cos x}\sqrt [3]{{x}^{2}+1}-{e}^{\cos x}}$
$=\lim _{x\rightarrow 0}\dfrac {e\cos x(-\sin x)}{\cos x-\sin x+\frac {1}{3}({x}^{2}+1){e}^{\cos x}(-\sin x)}$
$=\lim _{x\rightarrow 0}\dfrac {e\cos x(-\sin x)}{-\sin x-\sin x-\frac {1}{3}{x}^{3}{e}^{\cos x}}$
$=\lim _{x\rightarrow 0}\dfrac {e\cos x}{-\frac {1}{3}{x}^{3}{e}^{\cos x}+\sin x+\sin x}$
$=\dfrac {1}{-\frac {1}{3}\times 0+0+0}$
$=\dfrac {1}{3}$
：$\lim _{x\rightarrow 0}\dfrac {e-{e}^{\cos x}}{\sqrt [3]{{x}^{2}+1}-1}$
$=\lim _{x\rightarrow 0}\dfrac {e({e}^{\cos x}-1)}{{e}^{\cos x}\sqrt [3]{{x}^{2}+1}-{e}^{\cos x}}$
$=\lim _{x\rightarrow 0}\dfrac {e\cos x(-\sin x)}{\cos x-\sin x+\frac {1}{3}({x}^{2}+1){e}^{\cos x}(-\sin x)}$
$=\lim _{x\rightarrow 0}\dfrac {e\cos x(-\sin x)}{-\sin x-\sin x-\frac {1}{3}{x}^{3}{e}^{\cos x}}$
$=\lim _{x\rightarrow 0}\dfrac {e\cos x}{-\frac {1}{3}{x}^{3}{e}^{\cos x}+\sin x+\sin x}$
$=\dfrac {1}{-\frac {1}{3}\times 0+0+0}$
$=\dfrac {1}{3}$