Answer
$\lim\limits_{x \to 0}{\frac{sin (5x)}{3x}} =\frac{5}{3}$
Work Step by Step
$\lim\limits_{x \to 0}{\frac{sin (5x)}{3x}} = \frac{5}{3}\times\lim\limits_{x \to 0}{(\frac{sin (5x)}{3x} \times \frac{3}{5})} =\frac{5}{3}\times\lim\limits_{x \to 0}{\frac{sin (5x)}{5x}}$
Let $t=5x$
$\frac{5}{3}\times\lim\limits_{t \to 0}{\frac{sin (t)}{t}} = \frac{5}{3}\times1=\frac{5}{3}$
