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 $x\to0$.

$$\frac{1}{\sin^2 x}=\frac{1}{(x-x^3/6+O(x^5))^2}=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/6+O(x^4))^2}\\=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/3+O(x^4))}=\frac{1}{x^2}(1+x^2/3+O(x^4))$$
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
$$\frac{\mathrm d^4 (x^2\sin^2x)}{\mathrm d x^4}=8(3-x^2)\cos(2x)-32x\sin(2x)$$
$$\frac{\mathrm d^4 (\sin^2x-x^2)}{\mathrm d x^4}=-8\cos(2x)$$
$$\lim_{x\to0}\frac{1}{f(x)}=\lim_{x\to0}\frac{8(3-x^2)\cos(2x)-32x\sin(2x)}{-8\cos(2x)}=-3$$
