# $\lim \limits_{x \to 0} \frac{1}{x^2}-\csc^2x$ using L'Hôpital's rule

This is from an exercise set on L'Hôpital's rule. I've never seen anything so stubborn and recalcitrant. Anybody have suggestions?

$\lim \limits_{x \to 0} \frac{1}{x^2}-\csc^2x$

As $x\to0$,
$$\frac{1}{x^2}-\frac{1}{\sin^2 x}=\frac{1}{x^2}-\frac{1}{(x-x^3/6+o(x^3))^2}=\frac{1}{x^2}-\frac{1}{x^2-x^4/3+o(x^4)}\\=\frac{1}{x^2}-\frac{1}{x^2}\left(\frac{1}{1-x^2/3+o(x^2)}\right)=\frac{1}{x^2}-\frac{1}{x^2}\left(1+x^2/3+o(x^2)\right)=-\frac{1}{3}+o(1)\to-\frac{1}{3}$$
HINT: Write this as $$\lim_{x\to 0}\left(\frac1{x^2} - \frac1{\sin^2x}\right)$$ and find a common denominator. Further hint: Using Taylor polynomials is way better than using L'Hôpital's rule 4 times.