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