calculate $f´(0)$ if: $f(x) = 0$ for $x = 0$, and $f(x) =x^2\sin(1/x)$ for $x \neq 0$ using the definition of the derivative

Question

calculate $f´(0)$ if: $f(x) = 0$ for $x = 0$, and $f(x) =x^2\sin(1/x)$ for $x \neq 0$ using the definition of the derivative

I know that f is continuous but Im not sure how to solve it by using the definition of the derivative. Thank you so much for those who respond.

Answer 1

$$ \frac {f(x)-f(0)}{x-0} =\frac {x^2\sin(1/x)-0}{x-0} = x\sin (1/x) $$

$$ lim_{x \to 0} x\sin (1/x) =0$$

$$ f'(0)=0$$

Answer 2

In order to evaluate the limit Robert was talking about think about the squeeze theorem. As

\begin{align} |\sin(\frac{1}{x})| \leq 1 \end{align}

you know that

\begin{align} 0 \leq |x\cdot \sin(\frac{1}{x})| \leq |x| \end{align}

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