# Is f differentiable at x=0?

#### evinda

##### Well-known member
MHB Site Helper
Hi!!! 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? Last edited:

#### ZaidAlyafey

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

#### evinda

##### Well-known member
MHB Site Helper
Is this $$\displaystyle g(x) \sin (x)$$ ?
No,it is $g(x)sin\frac{1}{x}$ !!! I had some typos.I edited my post right now!! Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
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? Yes it is. #### evinda

##### Well-known member
MHB Site Helper
Yes it is. Nice Thank you very much!!!