# How to find the limit 5(1-cos2x)/sin2x as x approach 0 from the left ?

Mar 18, 2017

${\lim}_{x \to {0}^{-}} \frac{5 \left(1 - \cos 2 x\right)}{\sin 2 x} = 0$

#### Explanation:

To evaluate:

${\lim}_{x \to {0}^{-}} \frac{5 \left(1 - \cos 2 x\right)}{\sin 2 x}$

we can divide numerator and denominator by $2 x$ and then substitute $t = 2 x$:

${\lim}_{x \to {0}^{-}} \frac{5 \left(1 - \cos 2 x\right)}{\sin 2 x} = 5 {\lim}_{x \to {0}^{-}} \frac{\frac{1 - \cos 2 x}{2 x}}{\frac{\sin 2 x}{2 x}} = 5 {\lim}_{t \to {0}^{-}} \frac{\frac{1 - \cos t}{t}}{\frac{\sin t}{t}}$

Now consider the well known limits:

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

${\lim}_{x \to 0} \frac{1 - \cos x}{x} = 0$

and we have:

${\lim}_{x \to {0}^{-}} \frac{5 \left(1 - \cos 2 x\right)}{\sin 2 x} = \frac{5 \cdot 0}{1} = 0$

graph{(5(1-cos(2x)))/sin(2x) [-10, 10, -5, 5]}

Mar 18, 2017

$0.$

#### Explanation:

We have, $1 - \cos 2 x = 2 {\sin}^{2} x , \mathmr{and} , \sin 2 x = 2 \sin x \cos x .$

$\therefore \text{ The Reqd. Limit=} {\lim}_{x \to 0} \frac{5 \left(2 {\sin}^{2} x\right)}{2 \sin x \cos x}$

$= 5 \left\{{\lim}_{x \to 0} \tan x\right\}$

$= 5 \left(\tan 0\right)$

$= 0.$

Since the Limit exists, we have,

${\lim}_{x \to 0 -} \frac{5 \left(1 - \cos 2 x\right)}{\sin 2 x} = 0 = {\lim}_{x \to 0 +} \frac{5 \left(1 - \cos 2 x\right)}{\sin 2 x} , \text{ too.}$

Enjoy Maths.!