Question

If $f\left(x\right)=x\cos \left(\frac{1}{x^2}\right),x\ne 0,f\left(0\right)=1$, then $f\left(x\right)$ at $x=0$ is

• A. left continuous
• B. discontinuous
• C. right continuous
• D. continuous

Answer 1

The correct option is B discontinuous

$f\left(x\right)=x\cos \frac{1}{x^2}$

we know

$-x\le x\cos \frac{1}{x^2}\le x$

$\underset{x\to 0}{lim}-x\le \underset{x\to 0}{lim}x\cos \frac{1}{x^2}\le \underset{x\to 0}{lim}x$

$\therefore$ By sandwitch theorem

$\underset{x\to 0}{lim}x\cos \frac{1}{x^2}=0$

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

Hence, $f\left(x\right)$ is Discontinious