Limits, Continuity & Differentiability
Differentiability at a Point — x²sin(1/x) type
nta_pyq_2023_jan
Grade None
Question:
Let $f(x)=\begin{cases}x^2\sin\left(\dfrac{1}{x}\right), & x\neq0\\0, & x=0\end{cases}$. Then at $x=0$:
f is continuous but not differentiable
f is continuous but f' is not continuous
f and f' both are continuous
f' is continuous but not differentiable
Step-by-Step Solution
Key Concept: Continuity: $f(0^+)=\lim_{h\to0}h^2\sin(1/h)=0=f(0)$, so $f$ is continuous. Differentiability: $f'(0^+)=\lim_{h\to0}\frac{h^2\sin(1/h)}{h}=0$, so $f$ is differentiable at 0 with $f'(0)=0$.
$f$ is differentiable at 0 but $f'$ is not continuous at 0.
Correct Answer: 2