Question

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

Answer 1

Here, f(x) =$\{\begin{array}{c}{x}^{2}sin\frac{1}{x},ifx\ne 0\\ o,ifx=0\end{array}$

LHL = $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\left(x^2\right)sin\frac{1}{x}$,Putting x=a-h as $x\to 0^{-}$ when $h\to 0$

$\therefore \underset{x\to 0}{lim}(0-h{)}^{2}sin\frac{1}{(0-h)}=\underset{h\to 0}{lim}(-h^{2}sin\frac{1}{h})$ $\therefore sin(-0)=-sin0)$

=$-0xsin\left(\infty \right)=0(\therefore -1\le sinx\le 1,\forall xϵR)$

RHL = $\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\left(x^2\right)\frac{1}{x}$

Putting x=a-h as $x\to a^{+}$ when $h\to 0$

$\therefore \underset{x\to 0}{lim}(0-h{)}^{2}sin\frac{1}{(0+h)}=\underset{h\to 0}{lim}{h}^{2}sin\frac{1}{h}=0(\infty )$ $\therefore -1\le sinx\le 1,\forall xϵR)$

$\therefore$ LHL=RHl = f(0). Thus, f(x) is continuous at x=0