Continuity and Differentiability
Differentiability of Oscillating Functions
Premium Question
Grade 12
Question:
$f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f'(0)$ is:
(1) $0$
(2) $1$
(3) $-1$
(4) does not exist
Step-by-Step Solution
Key Concept: $f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h)$. Since $|h \sin(1/h)| \leq |h| \to 0$, the limit is 0.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)