Question

Let $f(x)=\begin{cases}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\\ 0 & , x=0\end{cases}$Then at $x=0$

• A. $f$ is continuous but not differentiable
• B. $f$ is continuous but $f^{\prime}$ is not continuous
• C. $f^{\prime}$ is continuous but not differentiable
• D. $f$ and $f^{\prime}$ both are continuous

Answer 1

Answer: (B)

$f$ is continuous but $f^{\prime}$ is not continuous

Answer 2

Continuity of f(x):f(0+)=h2⋅sinh1​=0
f(0−)=(−h)2⋅sin(h−1​)=0
f(0)=0
f(x) is continuous
f′(0+)=h→0lim​hf(0+h)−f(0)​=hh2⋅sin(h1​)−0​=0
f′(0−)=h→0lim​−hf(0−h)−f(0)​=−hh2⋅sin(−h1​)−0​=0
f(x) is differentiable.
f′(x)=2x⋅sin(x1​)+x2⋅cos(x1​)⋅x2−1​
f′(x)={2x⋅sin(x1​)−cos(x1​),0,​x=0x=0​
⇒f′(x) is not continuous (as cos(x1​) is highly oscillating at x=0 )