# solve the following limit without L'Hospital rule or series: <munder> <mo movablelimits="t

solve the following limit without L'Hospital rule or series:
$\underset{x\to 0}{lim}\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}$
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Johnathan Morse
First you want to show that
$\underset{y\to 0}{lim}\frac{\mathrm{arctan}y}{y}=1$
To do this, I would let $\theta =\mathrm{arctan}y$. Then $\theta \to 0$ as $y\to 0$, so
$\underset{y\to 0}{lim}\frac{\mathrm{arctan}y}{y}=\underset{\theta \to 0}{lim}\frac{\theta }{\mathrm{tan}\theta }=\underset{\theta \to 0}{lim}\frac{\theta }{\mathrm{sin}\theta }\cdot \mathrm{cos}\theta =1\cdot 1=1$
(the limit $\underset{\theta \to 0}{lim}\frac{\mathrm{sin}\theta }{\theta }=1$ is one of the basic limits you can usually invoke without proof).
Now
$\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}=\frac{\frac{\mathrm{arctan}2x}{2x}}{\frac{\mathrm{sin}3x}{3x}}\cdot \frac{2}{3}$