# What is the limit of (sin^2x)/(3x^2) as x approaches 0?

Apr 5, 2016

${\lim}_{x \to 0} {\sin}^{2} \frac{x}{3 {x}^{2}} = \frac{1}{3}$

#### Explanation:

Start with your favourite proof that ${\lim}_{x \to 0} \frac{\sin \left(x\right)}{x} = 1$

That might start with a geometric illustration that for small $x > 0$

$\sin \left(x\right) \le x \le \tan \left(x\right)$

Then divide through by $\sin \left(x\right)$ to get:

$1 \le \frac{x}{\sin} \left(x\right) \le \frac{1}{\cos} \left(x\right)$

Take reciprocals and reverse the inequality (since $\frac{1}{x}$ is monotonically decreasing for $x > 0$) to get:

$\cos \left(x\right) \le \sin \frac{x}{x} \le 1$

Then ${\lim}_{x \to 0 +} \cos \left(x\right) = 1$

So ${\lim}_{x \to 0 +} \left(\sin \frac{x}{x}\right) = 1$

Also $\sin \left(- x\right) = - \sin \left(x\right)$, so $\sin \frac{- x}{- x} = \sin \frac{x}{x}$

Hence ${\lim}_{x \to 0 -} \left(\sin \frac{x}{x}\right) = 1$ too.

$\textcolor{w h i t e}{}$
Whatever method we use to find ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$, it is not too difficult to deduce:

${\lim}_{x \to 0} {\sin}^{2} \frac{x}{3 {x}^{2}} = \left({\lim}_{x \to 0} \sin \frac{x}{x}\right) \cdot \left({\lim}_{x \to 0} \sin \frac{x}{x}\right) \cdot \frac{1}{3} = \frac{1}{3}$