Question

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

• A. $n ∈ (0, 1)$
• B. $n ∈ [1, ∞)$
• C. $n ∈ (− ∞, 0)$
• D. $n = 0$

Answer 1

Answer: (A)

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