# How do you evaluate the limit sinx/tanx as x approaches 0?

Sep 21, 2016

$L {t}_{x \to 0} \sin \frac{x}{\tan} x = 1$

#### Explanation:

$L {t}_{x \to 0} \sin \frac{x}{\tan} x$

= $L {t}_{x \to 0} \sin \frac{x}{\sin \frac{x}{\cos} x}$

= $L {t}_{x \to 0} \sin x \times \cos \frac{x}{\sin} x$

= $L {t}_{x \to 0} \cos x$

= $\cos 0$

= $1$

Sep 21, 2016

${\lim}_{x \rightarrow 0} \sin \frac{x}{\tan} x = 1$.

#### Explanation:

We use the Standard Form of Limit :

${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1 , \theta \in \mathbb{R}$.

From the above, we have,

${\lim}_{\theta \rightarrow 0} \tan \frac{\theta}{\theta}$

$= {\lim}_{\theta \rightarrow 0} \left\{\sin \frac{\theta}{\theta} \cdot \frac{1}{\cos} \theta\right\}$

$= \left\{{\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta}\right\} \cdot \left\{{\lim}_{\theta \rightarrow 0} \frac{1}{\cos} \theta\right\}$

$= 1 \cdot \left(\frac{1}{\cos} 0\right)$

$\therefore {\lim}_{\theta \rightarrow 0} \tan \frac{\theta}{\theta} = 1.$

Now, ${\lim}_{x \rightarrow 0} \sin \frac{x}{\tan} x$

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

$= \left\{{\lim}_{x \rightarrow 0} \sin \frac{x}{x}\right\} \left\{{\lim}_{x \rightarrow 0} \frac{x}{\tan} x\right\}$

$= 1 \cdot 1$

$\therefore {\lim}_{x \rightarrow 0} \sin \frac{x}{\tan} x = 1$.