Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin 3x}{x},& \text{when}x\ne 0\\ 1,& \text{when}x=0\end{array}$ Find whether $f\left(x\right)$ is continuous at $x=0.$

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin 3x}{x},& \text{if}x\ne 0\\ 1,& \text{if}x=0\end{array}$

LHL at $=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}f(-h)$

$\therefore \underset{h\to 0}{lim}\frac{\sin (-3h)}{-h}=\underset{h\to 0}{lim}\frac{-\sin 3h}{-h}=\underset{h\to 0}{lim}\frac{3\sin 3h}{3h}=3\underset{h\to 0}{lim}\frac{\sin 3h}{3h}=3\times 1=3$

RHL at $=0$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}f\left(h\right)$

$\therefore \underset{h\to 0}{lim}\frac{\sin \left(3h\right)}{h}=\underset{h\to 0}{lim}\frac{3\sin 3h}{3h}=3\underset{h\to 0}{lim}\frac{\sin 3h}{3h}=3\times 1=3$

And $f\left(0\right)=1$

$\because \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)\ne f\left(0\right)$

Hence $f\left(x\right)$ is discontinuous at $x=0$.