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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?

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  • $\begingroup$ @Arthur I tried to use the definition of derivative. didn't get to anything. $\endgroup$ – gshrf Jan 2 '18 at 7:26
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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 $$\color{red}{f'(0) = \lim_{x\to0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to0}\frac{g(x)\cos(1/x^3)}{x} =0}$$

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  • $\begingroup$ Thanks for the edit and answer! What was the use of the given value of g''? @GuyFsone $\endgroup$ – gshrf Jan 2 '18 at 8:26
  • $\begingroup$ read your exercise sheet it might be useful int the sequels not here $\endgroup$ – Guy Fsone Jan 2 '18 at 8:32

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