calculus - Evaluate $mathop {lim }limits_{x to 0} left( {{1 over {{{sin }^2}x}} - {1 over {{x^2}}}} right)$

I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating

Is there a shortcut?
$$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$

Thanks in advance

Answer

Of course there is!

$$\sin x \sim x - \frac{x^3}{6}$$

$$\sin^2 x \sim x^2 - \frac{x^4}{3}$$

So $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$
$$= \lim_{x \to 0} \frac{x^2 - \sin^2 x}{x^2 \cdot \sin^2 x} = \lim_{x \to 0} \frac{\frac{x^4}{3}}{x^4 - \frac{x^6}{3}} = \frac{1}{3}$$

(cause also $x^4 \pm x^6 \sim x^4$ if $x \to 0$)