Is $g$ differentiable in $0$?

Asked Nov 23 '18 at 14:48

Active Nov 23 '18 at 15:29

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Let $g$ be a continuous function on $\mathbb{R}$ such that: $\displaystyle\lim_{x \to 0} \:\dfrac{g(2x)-g(x)}{x}=l \in \mathbb{R}.$

I am supposed to prove that $g$ is differentiable at $0$ and $g'(0)=l$

Here is what I did: $$\displaystyle\begin{align} \lim_{x \to 0} \:\dfrac{g(2x)-g(x)}{x}&=\lim_{x \to 0}\:\frac{2(g(2x)-g(0))}{2x}-\dfrac{g(x)-g(0)}{x}\\&=\lim_{X \to 0}\:2\:\dfrac{g(X)-g(0)}{X} - \lim_{x \to 0}\:\dfrac{g(x)-g(0)}{x}\\ &=\lim_{x \to 0}\:\dfrac{g(x)-g(0)}{x}\\ &=l \end{align}$$ Since $\displaystyle\lim_{x \to 0} \:\dfrac{g(x)-g(0)}{x} = l$, then g is differentiable in $0$ and $g'(0)=l$

Is this valid? I'd appreciate any suggestion thanks.

edited Nov 23 '18 at 15:29

Riccardo Ceccon

1,538

asked Nov 23 '18 at 14:48

Idriss Mo

2

How do you know $\lim\limits_{x\to 0}\frac{g(x)-g(0)}{x-0}$ exists? – Anupam Nov 23 '18 at 14:52
@Anupam You are right I didn't see that, I'll have to think about it more, thanks btw. – Idriss Mo Nov 23 '18 at 14:59
This is wrong. Take for instance $g(x)=x$. Then the limit $l$ is equal to $\frac12$, but $g'(0)=1$. Are you sure you have copied the question correctly? – TonyK Nov 23 '18 at 15:00
@TonyK wait I'll fix it. – Idriss Mo Nov 23 '18 at 15:03
@TonyK: What if $g(x)=1$? – Yadati Kiran Nov 23 '18 at 15:04
@TonyK I think it's correct now. – Idriss Mo Nov 23 '18 at 15:06
I have added a simple answer to this question to the slightly more general predecessor question https://math.stackexchange.com/questions/1864058/prove-that-f0-exists-and-f0-b-a-1/3011677#3011677 . – Christian Blatter Nov 24 '18 at 15:21

Is $g$ differentiable in $0$?

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