Answer
$\lim\limits_{x \to 0}{\frac{sin (3x)}{x}} = 3$
Work Step by Step
$\lim\limits_{x \to 0}{\frac{sin (3x)}{x}} = 3\times\lim\limits_{x \to 0}{(\frac{sin (3x)}{x} \times \frac{1}{3})} =3\times\lim\limits_{x \to 0}{\frac{sin (3x)}{3x}}$
Let $t=3x$
$3\times\lim\limits_{t \to 0}{\frac{sin (t)}{t}} = 3\times1=3$
