# Given (sin(-2x)) / x how do you find the limit as x approaches 0?

Jul 8, 2016

-2

#### Explanation:

one way

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

let u = -2x

$= {\lim}_{u \to 0} \frac{\sin u}{- \frac{u}{2}}$

$= - 2 \textcolor{b l u e}{{\lim}_{u \to 0} \frac{\sin u}{u}}$

$= - 2 \cdot 1 = - 2$ as the term in blue is a well known limit often proved using squeeze theorem

OR

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

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

then using the fact that $\sin 2 \psi = 2 \sin \psi \cos \psi$
$= - {\lim}_{x \to 0} \frac{2 \sin x \cos x}{x}$

$= - 2 {\lim}_{x \to 0} \frac{\sin x \cos x}{x}$

then lifting $\cos x$ out as it is continuous throught the limit and $\cos 0 = 1$

$= - 2 \cos 0 \textcolor{g r e e n}{{\lim}_{x \to 0} \frac{\sin x}{x}}$

the term in green is indeterminate but as stated it is also a well known limit

you cannot use LHopital to prove this limit.