<s>[INST] How to calculate $\lim \limits_{x \to 0} \frac{x^2 \sin^2x}{x^2-\sin^2x}$ with $\lim \limits_{x \to 0} \frac{\sin x}{x}=1$? How to calculate
$$\lim \limits_{x \to 0} \frac{x^2 \sin^2x}{x^2-\sin^2x}$$ 
with
$$\lim \limits_{x \to 0} \frac{\sin x}{x}=1?$$
Yes I know the question has been asked, the answer is $3$, L'Hospital or Taylor series works and they are neat. However the question I meet is that I'm required to do it with the given limit. It may be dull algebra work; however I even don't know how to start with. Any idea will be wonderful.
 [/INST] We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{x^{2}\sin^{2}x}{x^{2} - \sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{x^{4}}{x^{2} - \sin^{2}x}\cdot\frac{\sin^{2}x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{x^{4}}{x^{2} - \sin^{2}x}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{x}{x + \sin x}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \lim_{x \to 0}\dfrac{1}{1 + \dfrac{\sin x}{x}}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \lim_{x \to 0}\frac{1}{1 + 1}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{x^{3}}{x - \sin x}\notag
\end{align}
Its difficult to do the last limit without Taylor's series or L'Hospital's Rule. The limit is famous and evaluates to $6$ so that the final answer is $3$.
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