Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit. 
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I do? 

Using Taylor series
$$\lim_{x\rightarrow0} \frac{\sin^2 x - x^2}{x^2\sin^2 x} = \lim_{x\rightarrow0} \frac{\Big(x - \frac{x^3}{6} + O(x^5)\Big)^2 - x^2}{x^2\Big(x - \frac{x^3}{6} + O(x^5)\Big)^2} = \lim_{x\rightarrow 0} \frac{\Big(x^2 - \frac{x^4}{3} + O(x^6)\Big) - x^2}{x^2\Big(x^2 - \frac{x^4}{3} + O(x^6)\Big)} \\= \lim_{x\rightarrow 0} \frac{-\frac{x^4}{3} + O(x^6)}{x^4 + O(x^6)} = -\frac{1}{3}$$
