Question

Examine the continuity of the function $f\left(x\right)$ at $x=0$, where
$f\left(x\right)=x\sin$$\frac{1}{x},$ for $x\ne 0$
$=0,$ for $x=0$

Answer 1

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

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

$\therefore \underset{x\to 0^{-}}{lt}f\left(x\right)=f\left(0\right)=\underset{x\to 0^{+}}{Lt}f\left(x\right)$

$\therefore$ The given function is continuous.