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? 

Since you already received good answers and being myself in love with Taylor series for more than 55 years, let me add a small trick.
Considering the expression $$\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$$ you have, as already given in answers, for the numerator $$-\frac{x^4}{3}+\frac{2 x^6}{45}-\frac{x^8}{315}+O\left(x^9\right)$$ and for the denominator $$x^4-\frac{x^6}{3}+\frac{2 x^8}{45}+O\left(x^9\right)$$ The ratio gives you immediately the limit.
But, make one further step, using long division for example and you will arrive to $$\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}\approx-\frac{1}{3}-\frac{x^2}{15}$$ which gives you the limits and how it is approached.
Just as a curiosity, plot on the same graph the function and this last approximation for $-1 \leq x\leq 1$; you will be surprized to see how close the curves are over this quite large interval.
If you push the long division for one more term, you should get $$\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}\approx-\frac{1}{3}-\frac{x^2}{15}-\frac{2 x^4}{189}$$ and now the curves are almost identical.
