Question

Show that $f\left(x\right)=\frac{x\tan 2x}{\sin 3x\sin 5x}$ for $x\ne 0$.

Answer 1

$f\left(0\right)=\frac{2}{17}$ is discontinuous at $x=0$

$f\left(x\right)=\frac{x\tan 2x}{\sin 3x\sin 5x}$ as $x\to 0$

$=\frac{x\frac{\tan 2x}{2x}\times 2x}{\frac{\sin 3x}{3x}\times 3x\frac{\sin 5x}{5x}\times 5x}$

$=\frac{x\times 2x}{3x\times 5x}$

$=\frac{2}{15}\ne f\left(0\right)$

So,$f\left(x\right)$ is discontinuous at $x=0$