题目

lim _(xarrow 0)dfrac (sin 2x-2sin x)(x(sqrt [3]{1+{x)^2}-1)}.

$\lim_{x\rightarrow0}\frac{\sin2x-2\sin x}{x\left(\sqrt[3]{1+x^2}-1\right)}$ .

题目解答

答案

$\lim _{x \rightarrow 0} \frac{\sin 2 x-2 \sin x}{x\left(\sqrt[3]{1+x^2}-1\right)}$

$=\lim _{x \rightarrow 0} \frac{2 \sin x \cos x-2 \sin x}{x\left[\left(1+x^2\right)^{\frac{1}{3}}-1\right]}$

$=\lim _{x \rightarrow 0} \frac{2 \sin x(\cos x-1)}{x\left[\left(1+x^2\right)^{\frac{1}{3}}-1\right]}$

∵有等价无穷小，即：当 $f(x) \rightarrow 0$ 时， $\sin f(x)\sim f(x)$ ， $\cos f\left(x\right)-1\sim-\frac{1}{2}f^2\left(x\right)$ ， $\left[1+f\left(x\right)\right]^a-1\sim af\left(x\right)$

∴ $\lim _{x \rightarrow 0} \frac{2 \sin x(\cos x-1)}{x\left[\left(1+x^2\right)^{\frac{1}{3}}-1\right]}$

$=\lim _{x \rightarrow 0} \frac{2 x \cdot\left(-\frac{1}{2} x^2\right)}{x \cdot\left(\frac{1}{3} x^2\right)}$

$=\lim _{x \rightarrow 0} \frac{-x^3}{\frac{1}{3} x^3}$

$=-3$

故本题答案为 $-3$ .

解析

步骤 1：化简分子
$\sin 2x - 2\sin x = 2\sin x\cos x - 2\sin x = 2\sin x(\cos x - 1)$
步骤 2：化简分母
$x(\sqrt[3]{1+x^2} - 1) = x\left((1+x^2)^{\frac{1}{3}} - 1\right)$
步骤 3：应用等价无穷小
当$x\rightarrow 0$时，$\sin x \sim x$，$\cos x - 1 \sim -\frac{1}{2}x^2$，$(1+x^2)^{\frac{1}{3}} - 1 \sim \frac{1}{3}x^2$
步骤 4：代入等价无穷小
$\lim _{x\rightarrow 0}\dfrac {2\sin x(\cos x-1)}{x[ {(1+{x}^{2})}^{\dfrac {1}{3}}-1] }$
$=\lim _{x\rightarrow 0}\dfrac {2x\cdot (-\dfrac {1}{2}{x}^{2})}{x\cdot (\dfrac {1}{3}{x}^{2})}$
$=\lim _{x\rightarrow 0}\dfrac {-{x}^{3}}{\dfrac {1}{3}{x}^{3}}$
步骤 5：计算极限
$=\lim _{x\rightarrow 0}\dfrac {-{x}^{3}}{\dfrac {1}{3}{x}^{3}} = -3$