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

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@TuckerRapu: The dependency is the other way around; this question is older than the one you've linked. –  user86418 Apr 9 at 10:46

2 Answers 2

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, $$g'(0) = 0$$

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Note that here is that the derivative is not continuous. Note that the derivative, $g'(0)=0$ and $g'(x)=2xsin(1/x)-cos(1/x)$. Let us now solve $g'(x)=2xsin(1/x)-cos(1/x)=1$ Let us now set $u=1/x$. Then the equation may be rewritten as $\frac{2sin(u)}{u}-cos(u)$. We therefore need to solve $2u=cot(u)$. If you look at the graph, you see that you have $u$-solutions as large as you want them to be. Thus, the $x$-solutions as small as you want. This means that you can find places arbitrarily close to 0 where the derivative is one. But the derivative at 0 is 0. Thus $g'$ is discontinuous at 0 –  Baby Dragon Nov 8 '12 at 5:55

We need to check differentiality only at $x=0$ since $x^2$ is differential everywhere and $\sin(\frac{1}{x})$ is differential everywhere except at $x=0$.

Now L.H.D.$= \lim_{h\to 0}\frac{h^2\sin(\frac{1}{h})-0}{h}=\lim_{h\to 0}h\sin(\frac{1}{h})=0$

Similarly, R.H.D.$=\lim_{h\to 0}\frac{(-h)^2\sin(\frac{1}{-h})-0}{-h}=\lim_{h\to 0}h\sin(\frac{1}{h})=0$

Since, L.H.D.=R.H.D., therefore, $g'(0)$ exists and equal to $0$.

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