How to compute the limit, $\lim_{x\to 0} (x^{-2}-\sin(x)^{-2})$? I've tried to combine the terms so as to compute the limit for $\frac{\sin(x)^{2}-x^2}{x^2\sin(x)^2}$. Then I tried to use L'Hopital's Rule to find derivatives for the denominator and nominator, but I ended up not being able to convert the denominator to a non-zero number (there's always an x involved so it becomes zero). There's probably another angle to approach this question, but I am not sure what.
 A: My approach: Since: $ \cot x = \frac{1}{x} - \frac{x}{3} + \mathcal{O} (x^2), x\to 0$. Hence:
$$\cot^2 x = \frac{1}{x^2} + \frac{x^2}{9} -\frac{2}{3} +\mathcal{O} (x), x\to 0$$
Therefore:
$$\lim_{x \to 0}\frac{1}{x^2} - \frac{1}{\sin^2 x} = \lim_{x \to 0} \frac{1}{x^2}-\left(1+ \frac{1}{x^2} + \frac{x^2}{9} -\frac{2}{3}+ \mathcal{O} (x)\right)=-\frac{1}{3}$$
A: After combining like terms, we can rearrange the expression like so
$$\frac{\sin^2x-x^2}{x^4}\cdot\frac{x^2}{\sin^2x} = \frac{\sin x-x}{x^3}\cdot\frac{\sin x + x}{x}\cdot\frac{x^2}{\sin^2x}$$
The limit on the left can be evaluated without L'Hopital by using the substitution $x = 3t$. The product of the limits is
$$-\frac{1}{6}\cdot 2 \cdot 1 = -\frac{1}{3}$$
A: The power-series approach is often simpler to apply that L'Hopital's rule when you don't know how many itterations of L'Hopital's you may need to get past an indeterminate case.
If you don't like powerseries....
The 2nd derivivate of the denominator will leave you with a term with no x coefficient.   But alas that term will be $\sin^2 x.$  More differentiation required.  The 4th derivative will leave you with a non-zero term in the denominator.
Making this sort of inspection will give you some idea whether you need to keep taking derivatives.
The substitution $\sin^2 x = \frac 12 - \cos 2x$ might save you a little bit of work.
$\frac {\sin^2 x - x^2}{x^2\sin^2 x}\\
\frac {1-2x^2 - \cos 2x}{x^2(1-\cos 2x)}\\
\frac {1-2x^2 - (1-\frac {(2x)^2}{2} + \frac {(2x)^4}{4!} - O(x^6))}{x^2(1-(1-\frac {(2x)^2}{2} + \frac {(2x)^4}{4!} - O(x^6)))}\\
\lim_\limits{x\to 0}\frac {- \frac {(2x)^4}{4!} - O(x^6))}{x^2(\frac {(2x)^2}{2} - O(x^4))}\\
\frac {-1}{3}$
