Question

Determine if $f$ defined by
$f\left(x\right)=$ $\{\begin{array}{c}{x}^2\sin \frac{1}{x},\text{if}x\ne 0\\ 0,\text{if}x=0\end{array}$
is a continuous function ?

Answer 1

$f\left(x\right)=\{\begin{array}{c}{x}^2\sin \frac{1}{x};x\ne 0\\ 0;x=0\end{array}$

At $x=0$

A function is continuous at $x=0$

if L.H.L$=$R.H.L$=f\left(0\right)$

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

L.H.L$={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}{x}^2\sin \frac{1}{x}$

Put $x=0-h$

As $x\to 0,0-h\to 0,h\to 0$

$={lim}_{h\to 0}{(0-h)}^2\sin \frac{1}{(0-h)}$

$={lim}_{h\to 0}{(-h)}^2\sin \frac{1}{(-h)}$

$={lim}_{h\to 0}{h}^2\sin \frac{1}{(-h)}$

$=-{lim}_{h\to 0}{h}^2\sin \frac{1}{h}$ as $\sin (-x)=-\sin x$

We know that $-1\le \sin \theta \le 1$

$\Rightarrow -1\le \sin \frac{1}{h}\le 1$

$\therefore \sin \frac{1}{h}$ is a finite value

Let $\sin \frac{1}{h}=k$

$=-{lim}_{h\to 0}{h}^{2}k$

Putting $h=0$

$=0\times k=0$

$\therefore$L.H.L$=0$

R.H.L$={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}{x}^2\sin \frac{1}{x}$

Put $x=0+h$

As $x\to 0,0+h\to 0,h\to 0$

$={lim}_{h\to 0}{(0+h)}^2\sin \frac{1}{(0+h)}$

$={lim}_{h\to 0}{\left(h\right)}^2\sin \frac{1}{\left(h\right)}$

$={lim}_{h\to 0}{h}^2\sin \frac{1}{\left(h\right)}$

$={lim}_{h\to 0}{h}^2\sin \frac{1}{h}$

We know that $-1\le \sin \theta \le 1$

$\Rightarrow -1\le \sin \frac{1}{h}\le 1$

$\therefore \sin \frac{1}{h}$ is a finite value

Let $\sin \frac{1}{h}=k$

$={lim}_{h\to 0}{h}^{2}k$

Putting $h=0$

$=0\times k=0$

$\therefore$R.H.L$=0$

and $f\left(0\right)=0$

Thus,L.H.L$=$R.H.L$=f\left(0\right)$

Hence $f\left(x\right)$ is continuous for all real value of $x$