Question

If function $f\left(x\right)=\{\begin{array}{c}x\sin \left(\frac{1}{x}\right);x\ne 0\\ a;x=0\end{array}$ is continuous at $x=0$, then the value of $a$ is

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

We have, $f\left(0\right)=a$
$\therefore \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}h\sin \frac{1}{h}=0$
and $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)$
$=\underset{h\to 0}{lim}(-h)\sin \frac{1}{-h}=0$
Since, $f\left(x\right)$ is continuous at $x=0$, we must have
$f\left(0\right)=\underset{x\to {0+}^{}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)\Rightarrow a=0$.