1. ## Differntiable problem

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!

2. 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.

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 $g(0)=g'(0)=0$
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.