Limit of $\frac{1}{x^2}-\frac{1}{\sin^2(x)}$ as $x$ approaches $0$ 
Evaluate
  $$\lim \limits_{x \rightarrow  0}\left(\frac{1}{x^2}-\frac{1}{\sin^2(x)}\right)$$

I tried to combine the fractions
$$\frac{1}{x^2}-\frac{1}{\sin^2(x)} = \frac{\sin^2(x)-x^2}{x^2\sin^2(x)}$$
and apply L'Hopitals which only made a mess.
I feel like there is a simpler way of doing this but I'm not quite sure what to do

Using the Taylor series for $\sin x$, we have that $\sin x = x - \dfrac{1}{6}x^3+O(x^5)$ for $x$ near $0$.
Thus, $\sin^2 x = \left(x - \dfrac{1}{6}x^3+O(x^5)\right)^2 = x^2 - \dfrac{1}{3}x^4 + O(x^6)$. 
Now, plug this into $ \dfrac{\sin^2 x - x^2}{x^2\sin^2 x}$ and see what you get. 
