Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$

Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:

$$\begin{align} I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\ = & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\ \implies 4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)\\ \implies 3I = & \lim_{x\to 0}\left\{\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right) -\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)\right\}\\ =&-\lim_{x\to 0} \frac{1-\cos^2 x}{\sin^2 x\cos^2 x} = -\lim_{x\to 0}\frac{1}{\cos^2 x} = -1\\ \implies I = & -\frac{1}{3} \end{align}$$

Well, you have $$\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) = \left(\frac{1}{x} - \frac{1}{\sin x}\right)\left(\frac{1}{x} + \frac{1}{\sin x}\right)$$ $$=\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)\left(1+ \frac{x}{\sin x}\right)$$ Since the limit of $\left(1+ \frac{x}{\sin x}\right)$ is $2$, your answer will be $$2\lim_{x \rightarrow 0}\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)$$ $$=2\lim_{x \rightarrow 0}\left(\frac{x}{\sin x}\right)\left(\frac{\sin x - x}{x^3}\right)$$ $$=2\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}$$ An elementary calculus way of showing this limit is $-{1 \over 6}$ can be found here.

So the overall answer is $-\frac{1}{3}$.