# How do you find the limit of (x-sin abs[x])/(3x^3)  as x approaches 0?

Jul 9, 2016

${\lim}_{x \to {0}^{-}} \frac{x - \sin \left\mid x \right\mid}{3 {x}^{3}} = \infty$

${\lim}_{x \to {0}^{+}} \frac{x - \sin \left\mid x \right\mid}{3 {x}^{3}} = \frac{1}{18}$

#### Explanation:

${\lim}_{x \to 0} \frac{x - \sin \left\mid x \right\mid}{3 {x}^{3}}$

that $\left\mid x \right\mid$ thing is a real pain so we can consider the different forms.

for x < 0, the limit becomes

${\lim}_{x \to {0}^{-}} \frac{x - \sin \left(- x\right)}{3 {x}^{3}}$

$= {\lim}_{x \to {0}^{-}} \frac{x + \sin x}{3 {x}^{3}} = \frac{0}{0}$ indeterminate so we use LHopital

$= {\lim}_{x \to {0}^{-}} \frac{1 + \cos x}{9 {x}^{2}}$

$\cos x$ is continuous at x = 0 so the limit is $\frac{2}{0} = \infty$

for x > 0, the limit becomes

${\lim}_{x \to {0}^{+}} \frac{x - \sin x}{3 {x}^{3}} = \frac{0}{0}$

indeterminate so again we use LHopital

$= {\lim}_{x \to {0}^{+}} \frac{1 - \cos x}{9 {x}^{2}} = \frac{0}{0}$

so we go again

$= {\lim}_{x \to {0}^{+}} \frac{\sin x}{18 x} = \frac{1}{18} {\lim}_{x \to {0}^{+}} \frac{\sin x}{x} = \frac{1}{18}$

as ${\lim}_{x \to 0} \frac{\sin x}{x} = 1$