Question

Let $f\left(x\right)=\{\begin{array}{c}{x}^n\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}$ , then f(x) 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, $\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)=0$

$\Rightarrow \underset{x\to 0}{lim}{x}^n\sin \left(\frac{1}{x}\right)=0\Rightarrow n>0$

$f\left(x\right)$ is differential at $x=0$ if $\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$ exists finitely

$\Rightarrow \underset{x\to 0}{lim}\frac{x^n\sin \left(\frac{1}{x}\right)-0}{x}$ exists finitely

$\Rightarrow x^{n-1}\sin \left(\frac{1}{x}\right)$ exists finitely.

$\Rightarrow n-1>0\Rightarrow n>1$

If $n\le 1$, then $\underset{x\to 0}{lim}{x}^{n-1}\sin \left(\frac{1}{x}\right)$ does not exists and hence, $f\left(x\right)$ is not differentiable at $x=0$

Hence, $f\left(x\right)$ is continuous but not differentiable at $x=0$ for $0<n\le 1$

i.e., $n\in (0,1]$