Answer
$\lim\limits_{x \to 0}{\frac{sin (7x)}{sin(3x)}}=\frac{7}{3}$
Work Step by Step
$\lim\limits_{x \to 0}{\frac{sin (7x)}{sin(3x)}} $
Multiply and divide by 7 and 3
$=\lim\limits_{x \to 0}{\frac{\frac{7sin (7x)}{7x}}{\frac{3sin(3x)}{3x}}}$
Let $t=7x$ in numerator and $u = 3x$ in denominator
$=\frac{7}{3}\frac{\lim\limits_{t \to 0}{\frac{sin(7x)}{7x}}}{\lim\limits_{u \to 0}{\frac{sin(3x)}{3x}}}$
$=\frac{7}{3}\times\frac{1}{1}=\frac{7}{3}$
