Question

Let $f\left(x\right)=\left[\begin{array}{cc}{x}^n\sin \frac{1}{x},& xî€=0\\ 0,& x=0\end{array}\right]$
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
$\underset{x→0}{\lim }f(x)=f(0)⇒\underset{x→0}{\lim }{x}^n\sin \frac{1}{x}=0,∀n>0$
$f(x)$ is differentiable at $x=0$, if ${\lim }_{x→0}\frac{f(x)−f(0)}{x−0}$ exists finitely.
$⇒\underset{x→0}{\lim }\frac{f(x)−f(0)}{x−0}$ exists finitely
$⇒\underset{x→0}{\lim }{x}^{n−1}\sin \frac{1}{x}$ exists finitely
$⇒n−1>0⇒n>1$
Hence, $f(x)$ is continuous but not differentiable at $x=0$, if $n∈(0,1)$.