# Thread: Differntiable problem

1. ## Differntiable problem

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

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 $\displaystyle g(\frac{1}{n})=0$ for all those large natural numbers. Hope that helps.

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

I really don't know how to start this one, please give me some hints, thanks!
What tbgh says shows that $\displaystyle g(0)=0$. Indeed, $\displaystyle g$ is continuous at 0 and $\displaystyle \frac{1}{n}\to_n 0$, hence $\displaystyle g(0)=\lim_n g\left(\frac{1}{n}\right)=0$.

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