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How to I prove this?? The questions asks "Observe this function is differentiable on $\mathbb{R}$."

eChung00
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  • HINT: the derivative is defined on all R except 0. On 0, calculate the limit by definition. – LeeNeverGup Nov 23 '13 at 21:50
  • @LeeNeverGup Thanks for your comment. But I think I get how to prove this is differentialble at 0. The problem is at all points except 0. – eChung00 Nov 23 '13 at 21:53
  • @eChung00: do you know that $\frac{1}{x}$, $\sin$, $x^2$ are all differentiable (for $\frac{1}{x}$, outside $0$)? – Najib Idrissi Nov 23 '13 at 21:55

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It suffices to show (using the definition of the derivative) that

$$\lim_{x\to0}x\sin\left(\frac1 x\right)=0$$ since the $\sin$ function is bounded.

Added The fact that $f$ is differentiable on $\mathbb R^*$ is clear so let's show that $f$ is differentiable at $0$ which's equivalent to show that $$\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}x\sin\left(\frac1 x\right)\ \text{exists}$$