Question

If the function $f\left(x\right)$ is defined by
$f\left(x\right)=x\sin \frac{1}{x}$ for $x\ne 0$
$=k$ for $x=0$
is continuous at $x=0$, then $k=$..........

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $0$

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

Given, $f\left(x\right)$ is continous at $x=0$

$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

$⟹\underset{x\to 0}{lim}x\sin \frac{1}{x}=k$

We know that, $-1\le \sin \left(\frac{1}{x}\right)\le 1$

$\underset{x\to 0}{lim}x\sin \left(\frac{1}{x}\right)\le \underset{x\to 0}{lim}x\left(1\right)=\underset{x\to 0}{lim}x=0$

$⟹\underset{x\to 0}{lim}x\sin \frac{1}{x}\le 0$

Also, $\underset{x\to 0}{lim}x\sin \left(\frac{1}{x}\right)\ge \underset{x\to 0}{lim}x(-1)=-\underset{x\to 0}{lim}x=0$

$⟹\underset{x\to 0}{lim}x\sin \frac{1}{x}\ge 0$

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

Thus we get $k=0$

Hence, option A is correct.