Find the limit of $\frac{1}{1-\cos(x)}-\frac{2}{x^2}$ as $x$ approaches $0$

Question

I need to find $$\lim_{x\to 0}\left(\frac{1}{1-\cos(x)}-\frac{2}{x^2}\right)$$ I already found it using Taylor series. However, I'm looking for a solution without Taylor series expansion or L'Hopital's rule because the problem was given in a calculus class at a point when only limits had been studied.

Answer 1

With the substitution $x=2z$, the limit becomes $$ \lim_{z\to0}\left(\frac{1}{\sin^2z}-\frac{1}{z^2}\right)= \frac{1}{2}\lim_{z\to0}\frac{z^2-\sin^2z}{z^2\sin^2z}= \frac{1}{2}\lim_{z\to0}\frac{z-\sin z}{z^3}\frac{z+\sin z}{z}\frac{z^2}{\sin^2z} $$ (see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).

The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967