# How to find these type of question easily, without using L-hopitals rule?

## Apr 2, 2018

$\pi$

#### Explanation:

$\sin \frac{\pi {\cos}^{2} x}{x} ^ 2 = \sin \frac{\pi - \pi {\sin}^{2}}{x} ^ 2 = \sin \frac{\pi {\sin}^{2} x}{x} ^ 2$

but for small $\left\mid x \right\mid$ we have sinx = x -x^3/(3!) + O(x^5) and also

${\sin}^{2} x = {x}^{2} + O \left({x}^{4}\right)$ hence

${\lim}_{x \to 0} \sin \frac{\pi {\cos}^{2} x}{x} ^ 2 = {\lim}_{x \to 0} \sin \frac{\pi \left({x}^{2} + O \left({x}^{4}\right)\right)}{x} ^ 2 = \pi {\lim}_{x \to 0} \sin \frac{\pi \left({x}^{2} + O \left({x}^{4}\right)\right)}{\pi {x}^{2}} = \pi$

Apr 2, 2018

#### Explanation:

As Cesareo points out:

$\sin \frac{\pi {\cos}^{2} x}{x} ^ 2 = \sin \frac{\pi \left(1 - {\sin}^{2}\right)}{x} ^ 2$

$= \sin \frac{\pi - \pi {\sin}^{2}}{x} ^ 2$ $\text{ }$ (now use the difference formula)

$= \sin \frac{\pi {\sin}^{2} x}{x} ^ 2$

$= \sin \frac{\pi {\sin}^{2} x}{1} \frac{1}{x} ^ 2$

$= \sin \frac{\pi {\sin}^{2} x}{\pi {\sin}^{2} x} \frac{\pi {\sin}^{2} x}{x} ^ 2$

$= \sin \frac{\theta}{\theta} \pi {\left(\sin \frac{x}{x}\right)}^{2}$

At $x \rightarrow 0$, we also have $\theta \rightarrow 0$, so the limit is

$1 \left(\pi\right) {\left(1\right)}^{2} = \pi$