Question

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

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{\sin 3x}{x},\mathrm{when}x\ne 0\\ 1,\mathrm{when}x=0\end{array}\right.$

We observe
(LHL at x = 0) = $\underset{\mathrm{x}\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)$
= ​$\underset{h\to 0}{\lim }\frac{\sin \left(-3h\right)}{-h}=\underset{h\to 0}{\lim }\frac{-\sin \left(3h\right)}{-h}=\underset{h\to 0}{\lim }\frac{3\sin \left(3h\right)}{3h}=3\underset{h\to 0}{\lim }\frac{\sin \left(3h\right)}{3h}=3·1=3$

(RHL at x = 0) = $\underset{\mathrm{x}\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(h\right)$
= ​$\underset{h\to 0}{\lim }\frac{\sin 3h}{h}=\underset{h\to 0}{\lim }\frac{3\sin 3h}{3h}=3\underset{h\to 0}{\lim }\frac{\sin \left(3h\right)}{3h}=3·1=3$

Given:
$f\left(0\right)=1$

It is known that for a function $f\left(x\right)$ to be continuous at x = a,
$\underset{\mathrm{x}\to a^{-}}{\lim }f\left(x\right)=\underset{\mathrm{x}\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)$

But here,
$\underset{\mathrm{x}\to 0^{-}}{\lim }f\left(x\right)=\underset{\mathrm{x}\to 0^{+}}{\lim }f\left(x\right)\ne f\left(0\right)$

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