How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? $$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$
Find limit of $\dfrac1{f(x)}$ as $x\to0$.

You don't really need to work with $1/f(x)$; consider
$$
f(x)=
\frac{\sin^2x-x^2}{x^4}\frac{x^2}{\sin^2x}=
\frac{\sin x-x}{x^3}\frac{\sin x+x}{x}\frac{x^2}{\sin^2x}=
\frac{\sin x-x}{x^3}\left(\frac{\sin x}{x}+1\right)
\left(\frac{x}{\sin x}\right)^2
$$
Since the limits of the second and third factors are well known, we just need to consider
$$
\lim_{x\to0}\frac{\sin x-x}{x^3}=
\lim_{x\to0}\frac{\cos x-1}{3x^2}=
\lim_{x\to0}\frac{-\sin x}{6x}=-\frac{1}{6}
$$
Therefore
$$
\lim_{x\to 0}f(x)=-\frac{1}{6}\cdot 2\cdot 1^2=-\frac{1}{3}
$$
and so
$$
\lim_{x\to0}\frac{1}{f(x)}=\frac{1}{-1/3}=-3
$$
The limit above can be computed also with a Taylor expansion:
$$
\lim_{x\to0}\frac{\sin x-x}{x^3}=
\lim_{x\to0}\frac{x-x^3/6+o(x^3)-x}{x^3}=-\frac{1}{6}
$$
Some knowledge of “basic” limits such as this one allows to use l'Hôpital more efficiently.
