Question

Find the set of values of $m$ for which $f\left(x\right)=\{\begin{array}{c}{x}^m\sin \left(\frac{1}{x}\right)x>0\\ 0x=0\end{array}$

is discontinuous at $x=0$

• A. $m\in (-\infty ,0)$
• B. $m\in (0,\infty )$
• C. $m\in (-\infty ,0)$
• D. $m\in (0,\infty )$

Answer 1

The correct option is C $m\in (-\infty ,0)$

Given, $f\left(x\right)=\{\begin{array}{c}{x}^m\sin \left(\frac{1}{x}\right)x>0\\ 0x=0\end{array}$

$\underset{x\to 0^{+}}{lim}{x}^{m}sin\left(\frac{1}{x}\right)=\underset{x\to 0+h}{lim}{h}^{m}sin\left(\frac{1}{h}\right)$

For discontinuity

$f\left(0^{+}\right)\ne f\left(0\right)⟹h^{m}sin\left(\frac{1}{h}\right)\ne 0$

Case-$1$:

If $m>0$,

then

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

Case-$2$:

If $m<0$,

then

$\underset{x\to 0}{lim}f\left(0^{+}\right)\ne 0$

Hence the set of values of $m$ for which $f\left(x\right)$ is discontinuous at $x=0$ is $m\in (-\infty ,0)$