Trigonometric limit without L'Hospital rule

Question

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!

Answer 1

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?

Answer 2

$$\lim_{x\rightarrow 0}\frac{\arctan{2x}}{\sin{3x}} = \lim_{x\rightarrow 0}\frac{\frac{\arctan{2x}}{2x}}{\frac{\sin{3x}}{3x}}\times \frac{2x}{3x} = \frac{2}{3}$$