Evaluating $ \lim_{x\to 0} ({\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}})$ without L'hopital's rule? $$ \lim_{x\to 0} ({\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}})$$
I know I can so some L'hopital's rule and get  the answer, but is there another way, more clever? Like with Taylor expansion? I tried it and got $\infty*0$

An alternative view is the following.
\begin{align}
\frac{1}{1- \cos x} - \frac{2}{x^{2}} &= \frac{1}{1 - \left(1 - \frac{x^{2}}{2!} - \frac{x^{4}}{4!} + \cdots \right)} - \frac{2}{x^{2}} \\
&= \frac{2}{x^{2}} \, \left[ \frac{1}{1 - \frac{2 \, x^{2}}{4!} + \frac{2 \, x^{4}}{6!} - \cdots } - 1 \right].
\end{align}
Now, by "long division" it is seen that
\begin{align}
\frac{1}{1 - \frac{2 \, x^{2}}{4!} + \frac{2 \, x^{4}}{6!} - \cdots } = 1 + \frac{2 \, x^{2}}{4!} + \frac{3 \, x^{4}}{6!} + \mathcal{O}(x^{6})
\end{align}
for which
\begin{align}
\frac{1}{1 - \cos x} - \frac{2}{x^{2}} &= \frac{2}{x^{2}} \, \left[ \left( 1 + \frac{2 \, x^{2}}{4!} + \frac{3 \, x^{4}}{6!} + \mathcal{O}(x^{6}) \right) - 1 \right] \\
&= \frac{1}{3!} \, \left[ 1 + \frac{x^{2}}{20} + \mathcal{O}(x^{4}) \right]. 
\end{align}
By taking the limit as $x \to 0$ leads to
\begin{align}
\lim_{x \to 0} \left\{ \frac{1}{1- \cos x} - \frac{2}{x^{2}} \right\} = \frac{1}{3!}.
\end{align}
