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? 

Outline: If we can write down the Taylor series for $\sin^2 x$, or just the first couple of terms, we will be finished. 
This can be done from $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ by ordinary squaring, treating the series for $\sin x$ casually as a "long" polynomial. Maybe it will feel better, however, to use the trigonometric identity
$$\cos 2x=1-2\sin^2 x, \quad\text{rewritten as}\quad \sin^2 x=\frac{1}{2}(1-\cos 2x).$$
Now from the series for $\cos t$ we get that the series for $\sin^2 x$ is
$$\frac{2x^2}{2!}-\frac{8x^4}{4!}+\frac{32x^6}{6!}-\cdots.$$
Now it's over. The leading term of $\sin^2 x-x^2$ is $-\frac{8x^4}{4!}$ and the leading term of $x^2\sin^2 x$ is $\frac{2x^4}{2!}$. Cancel. 
