$\lim_{x\rightarrow 0}$ $\frac{f(x)-f(cx)}{x}=b$ $\Rightarrow$ $f$ is differentiable at $0$

Question

Let $f:(-a,a)\rightarrow \mathbb{R}$, with $a>0$. If $f$ is continuous in $0$ and the limit

$$\lim_{x\rightarrow 0} \frac{f(x)-f(cx)}{x}$$

exists and is real, $0<c<1$, prove $f$ is differentiable at $0$.

My idea was this:

$f$ is continuous in $0 \Rightarrow$ $\lim_{x\rightarrow 0} f(x)=f(0)$.

$\lim_{x\rightarrow 0}$ $\frac{f(x)-f(cx)}{x}=b$

$\lim_{x\rightarrow 0}$ $\frac{f(cx)-f(x)}{x}=-b$

$c=0+c$

$\lim_{x\rightarrow 0}$ $\frac{f(cx+0)-f(x)}{x}=-b$

$\lim_{x\rightarrow 0}$ $\frac{f(c0+0)-f(0)}{x}=-b$

This is similar enough to

$f'(0)=\lim_{h\rightarrow 0}$ $\frac{f(h+0)-f(0)}{h}$

So it is valid that $f$ is differentiable at $0$. If not what did I go wrong with the argument?

Towards the end you replace some of the $x$s by $0$ and leave the one unchanged in the denominator. That is not valid. All the $x$s should approach $0$ at the same time. — Gary, Oct 14 '21 at 01:10
Also, what does "This is similar enough to" mean? You can't just say that if something is true, and it looks kind of similar to something else, then the second thing is also true. — Dan Velleman, Oct 14 '21 at 01:14
But change all $x$ for $0$ i have $\frac{f(0)-f(0)}{0}$ i can't understand how this is non-divergent. — Alexandre God, Oct 14 '21 at 01:14
Note that the restriction that $c < 1$ is necessary since the limit $\displaystyle\lim_{x\to 0}\frac{|x|-|-x|}{x} = 0$ but $|x|$ is not differentiable at $0$. — Cameron Williams, Oct 14 '21 at 01:41
This Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$ answers your question, but "what you did wrong" was covered by the comments. — Sarvesh Ravichandran Iyer, Oct 14 '21 at 02:17

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