Question

If function $f\left(x\right)=\{\begin{array}{c}{x}^m\sin \frac{1}{x};x\ne 0\\ 0;x=0\end{array}$
Where $m\in N$, then the least value of $m$ for which $f^{\prime }\left(x\right)$ is continuous at $x=0$ is

• A. 1
• B. 2
• C. 3
• D. none

Answer 1

The correct option is C 3

$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{h^{m}sin\frac{1}{h}}{h}$ must exist, i.e., m > 1. For m > 1,
$h^{\prime }\left(x\right)=\{\begin{array}{cc}m{x}^{m-1}sin\frac{1}{x}-x^{m-2}cos\frac{1}{x},& ifx\ne 0\\ 0,& ifx=0\end{array}$
Now, $\underset{h\to 0}{lim}h\left(x\right)=\underset{h\to 0}{lim}(m{h}^{m-1}sin\frac{1}{h}-h^{m-2}cos\frac{1}{2})$
Limit exists if m > 2.
Therefore, m $\in$ N or m $=$ 3.