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I am having some problems for solving the following limit without L'Hospital rule or series: $$\lim_{x\rightarrow 0}\frac{\arctan{2x}}{\sin{3x}}$$ Thank you very much for your help!

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  • $\begingroup$ Try a Taylor expansion at $0$. See what you get. $\endgroup$ – Teresa Lisbon Sep 15 '16 at 9:42
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First you want to show that $$ \lim_{y\to 0} \frac{\arctan y}{y} = 1 $$ To do this, I would let $\theta = \arctan y$. Then $\theta \to 0$ as $y \to 0$, so $$ \lim_{y\to 0} \frac{\arctan y}{y} = \lim_{\theta\to 0} \frac{\theta}{\tan \theta} = \lim_{\theta\to 0} \frac{\theta}{\sin \theta}\cdot \cos\theta = 1 \cdot 1 = 1 $$ (the limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$ is one of the basic limits you can usually invoke without proof).

Now $$ \frac{\arctan 2x}{\sin 3x} = \frac{\dfrac{\arctan 2x}{2x}}{\dfrac{\sin 3x}{3x}}\cdot \frac{2}{3} $$ Can you take it from there?

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  • $\begingroup$ Yes, thank you very much. I was not aware of the limit of arctan(x)/x. $\endgroup$ – user369151 Sep 15 '16 at 9:48
  • $\begingroup$ @user369151: You're welcome. $\arctan$ is just $\tan$ after reflecting the plane through the line $y=x$. $\endgroup$ – Matthew Leingang Sep 15 '16 at 9:51
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$$\lim_{x\rightarrow 0}\frac{\arctan{2x}}{\sin{3x}} = \lim_{x\rightarrow 0}\frac{\frac{\arctan{2x}}{2x}}{\frac{\sin{3x}}{3x}}\times \frac{2x}{3x} = \frac{2}{3}$$

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