Differntiable problem

**tttcomrader** · November 10th 2008, 09:41 AM

Suppose that the function $g: \mathbb {R} \rightarrow \mathbb {R}$ is differentiable at $x_0=0$ . Also, suppose that for each natural number n, $g( \frac {1}{n} ) = 0$ . Prove that $g(0)=g'(0)=0$

I really don't know how to start this one, please give me some hints, thanks!

**tbgh** · November 10th 2008, 10:13 AM

It's been too long since I did proofs so I don't have the language, but the key point to me seems to be that for any tiny distance around x=0, there are an infinite number of points such that g(x)=0. That's because of the second assumption that $g(\frac{1}{n})=0$ for all those large natural numbers. Hope that helps.

**Laurent** · November 10th 2008, 11:01 AM

Originally Posted by tttcomrader

Suppose that the function $g: \mathbb {R} \rightarrow \mathbb {R}$ is differentiable at $x_0=0$ . Also, suppose that for each natural number n, $g( \frac {1}{n} ) = 0$ . Prove that $g(0)=g'(0)=0$

I really don't know how to start this one, please give me some hints, thanks!

What tbgh says shows that $g(0)=0$ . Indeed, $g$ is continuous at 0 and $\frac{1}{n}\to_n 0$ , hence $g(0)=\lim_n g\left(\frac{1}{n}\right)=0$ .

Once we know $g(0)=0$ , we have, by definition, $g'(0)=\lim_{x\to 0, x\neq 0}\frac{g(x)}{x}$ , and this limit is assumed to exist because $g$ is differentiable. Then it is the same as for the first step.

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