calculate limit

Question

$\lim_{x\rightarrow 0} \left( \frac{1}{x^{2}} -\frac{1}{\sin^{2}x} \right)$

Answer 1

$\lim_{x\to 0}\left( \frac{1}{x^2} - \frac{1}{\sin^2 x}\right ) = \lim_{x\to 0}\left( \frac{\sin^2 x - x^2}{x^2 \sin^2 x}\right )$

Now when x --> 0, then $x^2 \approx 0$, and $\sin^2 x \approx 0$, so $x^2\sin^2 x \approx x^4$. So

$\lim_{x\to 0}\left( \frac{\sin^2 x - x^2}{x^2 \sin^2 x}\right ) = \lim_{x\to 0}\left( \frac{\sin^2 x - x^2}{x^4}\right )$

Now applying L'Hospital's rule, we get

$\lim_{x\to 0}\left( \frac{\sin^2 x - x^2}{x^4}\right ) = \lim_{x\to 0}\left( \frac{2\sin x\cos x - 2x}{4x^3}\right ) = \lim_{x\to 0}\left( \frac{\sin 2x - 2x}{4x^3}\right )$

Again applying L'Hospital's rule, we get

$\lim_{x\to 0}\left( \frac{\sin 2x - 2x}{4x^3}\right ) = \lim_{x\to 0}\left( \frac{2\cos 2x - 2}{12x^2}\right ) = \lim_{x\to 0}-2\left( \frac{1-\cos 2x}{12x^2}\right )$

$= \lim_{x\to 0}-2\left( \frac{2\sin^2 x}{12x^2}\right ) = \frac{-1}{3}$

Last equality comes by applying limit.