# How do you find the limit of (sin (2x)) / (sin (3x))  as x approaches 0?

Jul 24, 2016

${\lim}_{x \to 0} \sin \frac{2 x}{\sin} \left(3 x\right) = \frac{2}{3}$

#### Explanation:

This limit is indeterminate since direct substitution yields $\frac{0}{0}$, which means that we can apply L'Hospital's rule, which simply involves taking a derivative of the numerator and the denominator.

${\lim}_{x \to 0} \sin \frac{2 x}{\sin} \left(3 x\right) \to \frac{0}{0}$, so applying L'Hospital's rule:

${\lim}_{x \to 0} \frac{2 \cos \left(2 x\right)}{3 \cos \left(3 x\right)} = \frac{2}{3}$

Graph of $\sin \frac{2 x}{\sin} \left(3 x\right)$:

graph{sin(2x)/sin(3x) [-3.015, 3.142, -0.607, 2.471]}

Jul 24, 2016

2/3

#### Explanation:

${\lim}_{x \to 0} \sin \frac{2 x}{\sin} \left(3 x\right)$

${\lim}_{x \to 0} 2 \cdot \sin \frac{2 x}{2 x} \cdot \frac{1}{3} \frac{3 x}{\sin} \left(3 x\right)$

${\lim}_{x \to 0} 2 \cdot \sin \frac{2 x}{2 x} \cdot {\lim}_{x \to 0} \frac{1}{3} \frac{3 x}{\sin} \left(3 x\right)$

let $u = 2 x , v = 3 x$

${\lim}_{u \to 0} 2 \cdot \sin \frac{u}{u} \cdot {\lim}_{v \to 0} \frac{1}{3} \frac{v}{\sin} v$

$= 2 \cdot \frac{1}{3} = \frac{2}{3}$