Quickly evaluating this limit: $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right).$$
If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as
$$\frac{1}{x^2}-\frac{1}{\sin^2 x} = \frac{\sin^2 x - x^2}{x^2\sin^2 x} = \left(\frac{x^2}{\sin^2 x}\right)\left(\frac{\sin x + x}{x}\right)\left(\frac{\sin x - x}{x^{3}}\right),$$
where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.

Edit: Thanks to everyone's clear responses, I now understand:


*

*Taylor expansions of trig functions are immensely powerful in evaluating limits, in this case turning the problem into a limit of a rational function.

*If a factor of a limit exists and is nonzero, it can be factored out without affecting convergence of the product. Formally, suppose $\lim f(x)$ exists and $g(x)$ is a factor of $f(x)$.
Then if $\lim g(x)$ exists in $\mathbb{R}\backslash\{0\}$, $\lim f(x)/g(x)$ exists also. That is to say, one can be "opportunistic" about simplifying ones limits if a factor with a real nonzero limit is discovered.

$$
\frac {1}{x^2}-\frac{1}{\sin^2x} = \left(\frac {1}{x}+\frac{1}{\sin x}\right)\left(\frac {1}{x}-\frac{1}{\sin x}\right) = \frac{1}{x^2}\left(\frac {x}{x}+\frac{x}{\sin x}\right)\left(\frac {x}{x}-\frac{x}{\sin x}\right)
$$
so 
$$
\lim_{x\to 0}\left(\frac {1}{x^2}-\frac{1}{\sin^2x}\right) =2\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x}{\sin x}\right) = 2\lim_{x\to 0}\frac{x}{\sin x}\frac{1}{x^2}\left(\frac{\sin x}{x}-1\right) = 2\lim_{x\to 0}\frac{1}{x^2}\left(\frac{\sin x}{x}-1\right) = 2\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x^2}{3!}+o(x^4)-1\right) = -\frac 13
$$
