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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.

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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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  • $\begingroup$ I've been at this problem for a while... I'm beginning to doubt the possibility of doing it without Taylor's series or L'Hospital's rule (which your proof indirectly uses), and instead using only $\frac{\sin x}{x}$ $\endgroup$ Sep 22, 2015 at 3:45
  • $\begingroup$ @BrevanEllefsen: Check the linked answer in my post. It has some brilliant evaluations for the limit of $(x - \sin x)/x^{3}$. $\endgroup$
    – Paramanand Singh
    Sep 22, 2015 at 3:49
  • $\begingroup$ Wow, brilliant...so that's why I stuck for a night...Thanks you :) $\endgroup$
    – Asydot
    Sep 22, 2015 at 4:25

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