# Proving that $f'(0)=0$ with $f(x)={g(x)\cos(1/x^3) ~~for~~ x≠0 , 0~~ for ~~x=0}$

Assume that $g$ is a. differentiable function such that $g(0)=g'(0)=0$ , $g''(0)= -2$ we let $$f(x) = \begin{cases} g(x)\cos\big(\frac{1}{x^3}\big), & \text{if x \ne 0} \\[2ex] 0, & \text{if x = 0} \end{cases}$$

Show that $f'(0)=0$. can anyone help?

• @Arthur I tried to use the definition of derivative. didn't get to anything. – gshrf Jan 2 '18 at 7:26

Since $|\cos(a)|\le 1$ and $g'(0)=g(0)=0$ by definition of derivative we have
$$\bigg|\frac{g(x)\cos(1/x^3)}{x}\bigg|\le \bigg|\frac{g(x)}{x}\bigg| = \bigg|\frac{g(x)-g(0)}{x-0}\bigg|\to \left|g'(0)\right|=0~~~as ~~x\to0$$
Hence by definition derivative $$\color{red}{f'(0) = \lim_{x\to0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to0}\frac{g(x)\cos(1/x^3)}{x} =0}$$