Can the following $\infty - \infty$ limit be solved without using l'Hôpital rule?

Question

I know that some of the $\infty - \infty$ limits can be solved without using l'Hopital's rule. In such cases, you would usually be able to either rationalize (or derationalize if that is what its called) in the case of rational power expressions or you would take the lcm and cancel the common factors or simply apply squeeze theorem or Taylor's theorem. But I came across this question: $$ \lim_{x\to 0} \left(\,\frac{1}{x^{2}} - \frac{1}{(\sin x)^{2}}\right) $$

I am able to solve this using l'Hôpital's rule, and I get the limit at $0$ as $\frac{-1}{3}$ but I am not able to solve it using any other method. So, is there any alternative way to solve this limit?

PS- I changed $\frac{1}{3}$ to $\frac{-1}{3}$ as the latter is the correct solution. Thnx for everyone who edited my question to make it better and thnx for pointing out my mistakes.

Answer 1

$\lim\limits_{x\to0}\dfrac{\sin^2x-x^2}{x^2\sin^2x}=\lim\limits_{x\to0}\dfrac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}$

$=\lim\limits_{x\to0}\dfrac{(-\frac16x^3\cdots)(2x \cdots)}{x^4}\lim\limits_{x\to0}\dfrac x{\sin x}\lim\limits_{x\to0}\dfrac x{\sin x}$

$=-\frac13.$

Answer 2

Writing the functions involved as Taylor polynomials will often help. It does in this case. It’s simpler if you use $\frac{1}{f} - \frac{1}{g} = \frac{g - f}{fg}$.