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

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 $-\frac{1}{3}$.
