# Question #91010

May 3, 2016

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

#### Explanation:

Direct substitution of $x = 0$ shows that the initial limit results in a $\frac{0}{0}$ indeterminate form, and thus we can apply L'hopital's rule.

${\lim}_{x \to 0} \frac{\tan \left(x\right) - x}{x \sin \left(x\right)} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(\tan \left(x\right) - x\right)}{\frac{d}{\mathrm{dx}} x \sin \left(x\right)}$

$= {\lim}_{x \to 0} \frac{{\sec}^{2} \left(x\right) - 1}{x \cos \left(x\right) + \sin \left(x\right)}$

As this again results in a $\frac{0}{0}$ indeterminate form, we may apply L'hopital's rule once again:

$= {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left({\sec}^{2} \left(x\right) - 1\right)}{\frac{d}{\mathrm{dx}} \left(x \cos \left(x\right) + \sin \left(x\right)\right)}$

$= {\lim}_{x \to 0} \frac{2 {\sec}^{2} \left(x\right) \tan \left(x\right)}{2 \cos \left(x\right) - x \sin \left(x\right)}$

$= \frac{2 {\sec}^{2} \left(0\right) \tan \left(0\right)}{2 \cos \left(0\right) - 0 \sin \left(0\right)}$

$= \frac{2 \cdot 1 \cdot 0}{2 \cdot 1 - 0 \cdot 0}$

$= 0$