# What is the limit of sin(2x)/x^2 as x approaches 0?

Dec 11, 2014

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

by l'H$\hat{\text{o}}$pital's Rule (0/0),

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

Since

$\left\{\begin{matrix}{\lim}_{x \to {0}^{-}} \frac{\cos \left(2 x\right)}{x} = \frac{1}{0} ^ \left\{-\right\} = - \infty \\ {\lim}_{x \to {0}^{+}} \frac{\cos \left(2 x\right)}{x} = \frac{1}{0} ^ \left\{+\right\} = + \infty\end{matrix}\right.$,

${\lim}_{x \to 0} \frac{\cos \left(2 x\right)}{x}$ does not exist,

which means that

${\lim}_{x \to 0} \frac{\sin \left(2 x\right)}{{x}^{2}}$ does not exist.

Let us look at the graph of $y = \frac{\sin \left(2 x\right)}{x} ^ 2$.

The graph above indeed agrees with our conclusion.

I hope that this was helpful.