- Thread starter
- #1

- Apr 13, 2013

- 3,718

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?

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: