Answer
$\lim\limits_{x\to0}\dfrac{\sin{2x}}{\sin{3x}}=\dfrac{2}{3}.$
Work Step by Step
Using Theorem $1.9:$
$\lim\limits_{x\to0}\dfrac{\sin{2x}}{\sin{3x}}=\lim\limits_{x\to0}\dfrac{2\sin{2x}}{2x}\times\lim\limits_{x\to0}\dfrac{3x}{3\sin{3x}}
=\frac{2}{3}\lim\limits_{x\to0}\dfrac{\sin{2x}}{2x}\times\lim\limits_{x\to0}\dfrac{3x}{\sin{3x}}$
Let $y=2x$ and $z=3x$. Note that as $x\to0$, so does $y\to0$ and $z\to0$.
$=\frac{2}{3}\lim\limits_{y\to0}\dfrac{\sin{y}}{y}\times\lim\limits_{z\to0}\dfrac{z}{\sin{z}}=(\frac{2}{3})(1)(1)=\dfrac{2}{3}.$
