Question

Show that the function is continuous at $x=0$, if $f\left(x\right)=\frac{\sin 3x}{\tan 2x},x<0$

$=\frac{\log \left(1+3x\right)}{e^{2x}-1},x>0$

$=\frac{3}{2},x=0$

Answer 1

$f\left(x\right)=\frac{\sin 3x}{\tan 2x},x<0$

$=\frac{\log (1+3x)}{e^{2x}-1},x>0$

$=\frac{3}{2},x=0$

Checking continuity at $x=0$

$\Rightarrow \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\log (1+3(0+h\left)\right)}{e^{2(0+h)}-1}=\underset{h\to 0}{lim}\frac{\left(\frac{1}{3h+1}\right)\left(3\right)}{2e^{2h}}=\frac{3}{2}\to 1$ (By L' Hospital Rule)

$\Rightarrow \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\sin 3(0-h)}{\tan 2(0-h)}=\underset{h\to 0}{lim}\frac{(-\cos 3h)\left(3\right)}{-\left({\sec }^{2}2h\right).\left(2\right)}=\frac{3}{2}\to 2$ (By L' Hospital Rule)

$\Rightarrow f\left(0\right)=\frac{3}{2}\to 3$

From $1,2\&3$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)=\frac{3}{2}$

$\Rightarrow f\left(x\right)$ is continuous at $x=0$