Question

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

Answer 1

When $x=0$
Given: $f\left(x\right)=\{\begin{array}{cc}{x}^2\sin \frac{1}{x},& \text{if}x\ne 0\\ 0,& \text{if}x=0\end{array}$
If $f\left(x\right)$ is continuous at $x=0$ then
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$

L.H.L.
$=\underset{x\to 0^{-}}{lim}{x}^2{\sin }^2\frac{1}{x}=\underset{h\to 0}{lim}(0-h{)}^2\sin \frac{1}{(0-h)}$
$=\underset{h\to 0}{lim}(-h{)}^2\sin \frac{1}{(-h)}$
$=0\times$ ( oscillating between $-1$ to $1\left)\right)=0$

R.H.L.
$=\underset{x\to 0^{+}}{lim}{x}^2\sin \frac{1}{x}=\underset{h\to 0}{lim}(0+h{)}^2\sin \frac{1}{(0+h)}$
$={lim}_{h\to 0}{h}^2\sin \frac{1}{h}$
$=0\times$ (oscillating between $-1$ to $1)=0$

Finding $f\left(x\right)$ at $x=0$
$f\left(x\right)=0$ at $x=0$
$f\left(0\right)=0$
Hence, $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$

When $x\ne 0$
Given $f\left(x\right)=x^2\sin \frac{1}{x}$
Let us consider the function $p\left(x\right)=x^2,q\left(x\right)=\sin \frac{1}{x}$
$p\left(x\right)=x^2$ is polynomial function
So, $p\left(x\right)$ is continuous for $x\in R$
$q\left(x\right)=\sin \frac{1}{x}$ is sine function so, $q\left(x\right)$ is continuous for $x\ne 0$
By Algebra of continuous functions,
If $p\left(x\right)$ and $q\left(x\right)$ all are continuous for $x\ne 0$ then $f\left(x\right)=p\left(x\right).q\left(x\right)$ is continuous for $x\ne 0$.
$\therefore f\left(x\right)=\{\begin{array}{cc}{x}^2\sin \frac{1}{x},& \text{if}x\ne 0\\ 0,& \text{if}x=0\end{array}$ is everywhere continuous for all $x\in R$.