Question

Using the definition, show that the function.
$f\left(x\right)=x\sin \left(1/x\right)$ if $x\ne 0,0$ if $x=0$ is continuous at the point $x=0$

Answer 1

$f\left(x\right)=x\sin \left(\frac{1}{x}\right)forx\ne 0$

$andf\left(x\right)=0forx=0$

For a continuous function $\underset{x\to a}{L.H.L}=\underset{x\to a}{R.H.L.}=f\left(a\right)$

Let $L.H.L.$

$=\underset{x\to 0^{-}}{lim}x\sin \frac{1}{x}$

$=\underset{n\to 0}{lim}(0-4)\sin \left(\frac{1}{-n}\right)$

$\Rightarrow \underset{n\to 0}{lim}-n\sin \left(\frac{-1}{n}\right)$

$=\underset{n\to 0}{lim}n\sin \frac{1}{n}[\because \sin (-\theta )=-\sin \theta ]$

$=0\sin \left(\infty \right)$

$=0$

R.H.L,

$=\underset{x\to 0^{+}}{lim}x\sin \frac{1}{x}$

$=\underset{n\to 0}{lim}(0+n)\sin \frac{1}{0+n}$

$=\underset{n\to 0}{lim}n\sin \frac{1}{n}=0.\sin \infty$

$=0$

So, here

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

$0=0=0$

So function is continuous.