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How can I find $\lim \limits_{x\to 0} \frac{x \sin^2 x}{(\arctan 5x)^2}$ without using l'Hopital's rule?

I understand that I can simplify the numerator but what should I do with the denominator?

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  • $\begingroup$ the expression inside the limit is not clear. Is it $\sin^2 X$? $\endgroup$ – Babai Sep 30 '15 at 14:45
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Hint. Write $$ \frac{x \sin^2 x}{\arctan^2 5x} = \left(\frac{\arctan 5x}{5x}\right)^{-2} \cdot \left(\frac{\sin x}x\right)^2 \cdot \frac{x \cdot x^2}{25 x^2} $$ Now use the (hopefully known) limits of $\frac{\arctan x}{x}$ and $\frac{\sin x}x$ for $x \to 0$.

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