# How do you find the limit of tan^2x/x as x->0?

##### 1 Answer
Nov 24, 2016

${\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = 0$

#### Explanation:

${\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = {\lim}_{x \rightarrow 0} \frac{\tan x \tan x}{x}$

$\therefore {\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = {\lim}_{x \rightarrow 0} \sin \frac{x}{\cos} x \cdot \sin \frac{x}{\cos} x \cdot \frac{1}{x}$

$\therefore {\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = {\lim}_{x \rightarrow 0} \sin \frac{x}{x} \cdot \frac{1}{\cos} ^ 2 x \cdot \sin x$

$\therefore {\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = {\lim}_{x \rightarrow 0} \sin \frac{x}{x} \cdot {\lim}_{x \rightarrow 0} \frac{1}{\cos} ^ 2 x \cdot {\lim}_{x \rightarrow 0} \sin x$

$\therefore {\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = {\lim}_{x \rightarrow 0} \sin \frac{x}{x} \cdot {\left(\frac{1}{{\lim}_{x \rightarrow 0} {\cos}^{2} x}\right)}^{2} \cdot {\lim}_{x \rightarrow 0} \sin x$

${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 0$ is standard calculus limit

$\therefore {\lim}_{x \rightarrow 0} {\tan}^{2} \frac{x}{x} = 0 \cdot {1}^{2} \cdot 0 = 0$