Question

Let $f\left(x\right)=\left\{\begin{array}{c}{x}^n\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$. Then, $f\left(x\right)$ is continuous but not differentiable at $x=0$, if

• A. $n\in (0,1)$
• B. $n\in [1,\infty )$
• C. $n\in (-\infty ,0)$
• D. $n=0$

Answer 1

The correct option is A $n\in (0,1)$

Since, $f\left(x\right)$ is continuous at $x=0$, therefore
$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)\Rightarrow \underset{x\to 0}{lim}{x}^n\sin \frac{1}{x}=0,\forall n>0$
$f\left(x\right)$ is differentiable at $x=0$, if $\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$ exists finitely.
$\Rightarrow \underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$ exists finitely
$\Rightarrow \underset{x\to 0}{lim}{x}^{n-1}\sin \frac{1}{x}$ exists finitely
$\Rightarrow n-1>0$ $\Rightarrow$ $n>1$
Hence, $f\left(x\right)$ is continuous but not differentiable at $x=0$, if $n\in (0,1)$.