Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$ 
Show that the function $g(x) = x^2 \sin\left(\frac{1}{x}\right) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$.


HINT 1
The only problematic point is at $x=0$. (Why?)
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HINT 2 Use the definition of the derivative at a point i.e. $$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$$

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HINT 3 Recall that $\vert \sin(y) \vert$ is bounded by $1$.

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HINT 4 Use the above to sandwich the desired limit to give the answer.

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Solution $$\lim_{h \to 0} \dfrac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \dfrac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h)$$ Now note that $$\vert h \sin(1/h) \vert \leq \vert h \vert$$ Hence, we have that $$-\lim_{h \to 0} \vert h \vert \leq \lim_{h \to 0} h \sin(1/h) \leq \lim_{h \to 0} \vert h \vert$$ Hence, we get that $$\lim_{h \to 0} h \sin(1/h) = 0$$ Therefore, $$g'(0) = 0$$

