<s>[INST] 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? 
 [/INST] We have:
$$\lim_{x\to 0} \frac{\sin^2 x -x^2}{x^2\sin^2 x}$$
Since direct substitution yields $\frac{0}{0}$, we use L'Hospital's rule:
$$\lim_{x\to 0} \frac{2\sin x \cos x - 2x}{2x\sin^2 x + 2x^2\sin x \cos x }$$
Using it again:
$$\lim_{x\to 0} \frac{2\cos2x - 2}{x^2(2\cos^2 x-2\sin^2 x) + 2\sin^2 x +8x\sin x \cos x }$$
And again:
$$\lim_{x\to 0} \frac{-4\sin2x}{6x(2\cos^2 x-2\sin^2 x) + 12x\sin x \cos x -8x^2\sin x \cos x }$$
One more to go:
$$\lim_{x\to 0} \frac{-8\cos2x}{12(2\cos^2 x-2\sin^2 x) + 64x\sin x \cos x +x^2(8\cos^2 x-8\sin^2 x)} = \frac{-8}{24} = -\frac{1}{3}$$
Keep in mind this is a much longer method. L'Hospital's Rule is a simple yet sometimes very lengthy derivation of limits such as these. Any alternatives seen could most likely be easier to do.
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