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Is f differentiable at x=0?

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hi!!! :eek:
We know that $g$ is differentiable at $x=0$ with $g(0)=g'(0)=0$ and

$$f(x)=\left\{\begin{matrix}
g(x)sin\frac{1}{x} & ,x\neq 0\\
0& ,x=0
\end{matrix}\right. $$

Is $f$ differentiable at $x=0$.If yes,which is the value of $f'(0)$?

That's what I have tried:

If f is differentiable at $x=0$,then the limit $\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x \to 0}\frac{g(x)sin \frac{1}{x} }{x}$ exists.

As we know that g is differentiable at $x=0$,the limit $\lim_{x \to 0}\frac{g(x)-g(0)}{x-0}$ exists and is equal to $\lim_{x \to 0}\frac{g(x)}{x}=g'(0)=0$.
Also, $|\frac{g(x)sin\frac{1}{x}}{x} | \leq |\frac{g(x)}{x}|\Rightarrow -\frac{g(x)}{x}\leq \frac{g(x)sin\frac{1}{x}}{x}\leq \frac{g(x)}{x}$. We know that $\lim_{x \to 0}\frac{g(x)}{x}=0$ ,so from the squeeze theorem,the limit $\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}$ exists and is equal to 0.

Could you tell me if it is right? :eek:
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$$f(x)=\left\{\begin{matrix}
g(x)sin\frac{1}{x} & ,x\neq 0\\
0& ,x=0
\end{matrix}\right. $$
Is this \(\displaystyle g(x) \sin (x) \) ?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
We know that $\lim_{x \to 0}\frac{g(x)}{x}=0$ ,so from the squeeze theorem,the limit $\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}$ exists and is equal to 0.

Could you tell me if it is right? :eek:
Yes it is. :)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718