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 $x\to0$.
 A: As already noted:
$$\frac1{f(x)}=\frac{x^2\sin^2x}{\sin^2x-x^2}=\color{blue}{\frac {x\sin^2x}{\sin x-x}}\cdot\color{purple}{\frac x{\sin x+x}}\;\color{red}{(**)}$$
Now, using l'Hospital's rule:
$$\begin{align*}&\lim_{x\to0}\color{blue}{\frac{x\sin^2x}{\sin x-x}}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\sin^2x+x\sin2x}{\cos x-1}\stackrel{\text{l'H}}=\lim_{x\to0}2\,\frac{\sin2x+x\cos2x}{-\sin x}\stackrel{\text{l'H}}=\\{}\\
&=\lim_{x\to0}2\frac{4\cos2x-2x\sin2x}{-\cos x}=2\frac{4}{-1}=-6\;,\;\;\;\;\text{whereas}\\{}\\
&\lim_{x\to0}\color{purple}{\frac x{\sin x+x}}=\lim_{x\to0}\frac1{\frac{\sin x}x+1}=\frac1{1+1}=\frac12\end{align*}$$
Thus, we finally get the limit is
$$\color{red}{(**)}=-6\cdot\frac12=-3$$
A: 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.
A: Let's write explicitly 
$${1\over f(x)}={x^2\sin^2{x}\over \sin^2{x}-x^2}$$
Then use the Taylor expansion $\sin{x}=x-x^3/6+o(x^4)$. We can write
$${1\over f(x)}={x^4+o(x^4)\over {-x^4\over 3}+o(x^4)}=-3+o(1)$$
And the limit we're looking for is $-3$
A: $$\frac{1}{\sin^2 x}=\frac{1}{(x-x^3/6+O(x^5))^2}=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/6+O(x^4))^2}\\=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/3+O(x^4))}=\frac{1}{x^2}(1+x^2/3+O(x^4))$$
Hence
$$f(x)=\frac{1}{x^2}-\frac{1}{\sin^2 x}=-\frac13+O(x^2)$$
And
$$\frac{1}{f(x)}\underset{x\to 0}{\longrightarrow} -3$$

Without using Taylor expansion, you could apply L'Hôpital's rule to
$$\frac{1}{f(x)}=\frac{x^2\sin^2 x}{\sin^2 x-x^2}$$
However, you will need to differentiate four times, the numerator and the denominator (you get a $0/0$ form for lower derivatives). You will then get after simplification
$$\frac{\mathrm d^4 (x^2\sin^2x)}{\mathrm d x^4}=8(3-x^2)\cos(2x)-32x\sin(2x)$$
$$\frac{\mathrm d^4 (\sin^2x-x^2)}{\mathrm d x^4}=-8\cos(2x)$$
$$\lim_{x\to0}\frac{1}{f(x)}=\lim_{x\to0}\frac{8(3-x^2)\cos(2x)-32x\sin(2x)}{-8\cos(2x)}=-3$$
