Aspire Faculty ID #4303 · Topic: NIMCET 2017 · Just now
NIMCET 2017

The slope of the function \[ f(x) = \begin{cases} x^2 \sin\!\left(\dfrac{1}{x}\right), & \text{if } x \ne 0, \\[8pt] 0, & \text{if } x = 0 \end{cases} \]

Solution

$f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}$ 

$=\lim_{x\to0}\dfrac{x^2\sin(1/x)}{x}$ 

$=\lim_{x\to0}x\sin(\frac{1}{x})=0$

For $x\ne0$,

$f'(x)=2x\sin\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)$

Answer: $\boxed{f'(0)=0}$ ✅

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