Differentiability of $f(x)$ at $0$ provided that $\lim_{x\to0} [f(2x)-f(-x)]/x$ exists? Assume that $f:\mathbb{R}\to\mathbb{R}$ is continuous at $x=0$, and the limit
\begin{equation}
\lim_{x\to0}\frac{f(2x)-f(-x)}{x}
\end{equation}
exists. Is $f$ necessarily differentiable at 0?
I was able to deduce that if $f$ is continuous in a neighbordhood of $0$ and has an isolated zero at $x=0$, it must be differentiable (unless I made an error). I believe that this result holds if there exists some $\delta>0$ such that $f(x)$ does not change sign in the neiborhoods $-\delta<x<0$ and $0<x<\delta$.
But for the life of me I can't figure out a general proof (or a counterexample).
Any help would be greatly appreicated!
 A: Let $ \varepsilon >0 $, there exists some $ \eta >0 $ such that $ \left(\forall x\in\left(-\eta,\eta\right)\right),\ \left|f\left(2x\right)-f\left(-x\right)-\ell x\right|\leq\varepsilon\left|x\right| $.
Let $ n\in\mathbb{N}^{*} $, we have, for all $ x\in\left(-\eta,\eta\right) $ : \begin{aligned}\small\left|f\left(x\right)-f\left(0\right)-\frac{\ell}{3} x\right|&\small=\left|\sum_{k=0}^{n-1}{\left(f\left(\frac{\left(-1\right)^{k}}{2^{k}}x\right)-f\left(\frac{\left(-1\right)^{k+1}}{2^{k+1}}x\right)-\ell\frac{\left(-1\right)^{k}}{2^{k+1}}x\right)}+f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\sum_{k=0}^{n-1}{\left|f\left(\frac{\left(-1\right)^{k}}{2^{k}}x\right)-f\left(\frac{\left(-1\right)^{k+1}}{2^{k+1}}x\right)-\ell\frac{\left(-1\right)^{k}}{2^{k+1}}x\right|}+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\sum_{k=0}^{n-1}{\varepsilon\left|\frac{\left(-1\right)^{k}}{2^{k+1}}x\right|}+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\varepsilon\left(1-\frac{1}{2^{n}}\right)\left|x\right|+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\end{aligned}
Giving first some clarifications about what we did, we introduced first a telescopic sum combined with a geometric sum, then we applyied our hypothesis for $ y=\frac{\left(-1\right)^{k}}{2^{k+1}}x $, because whatever $ k $ is, $ y $ will always be in $ \left(-\eta,\eta\right) $, after that we simplified the geometric sum that we obtained to end up with the result.
How we've assumed that $ f'\left(0\right) $, if exists, must be $ \frac{\ell}{3} $ ? Well to get it, we must take look at the following, for $ x\in\left(-\eta,\eta\right) $ : $$ \frac{f\left(2x\right)-f\left(-x\right)}{x}=2\times\frac{f\left(2x\right)-f\left(0\right)}{2x}+\frac{f\left(-x\right)-f\left(0\right)}{-x} $$
Taking $ n $ to $ +\infty $, we end with : $$ \left|f\left(x\right)-f\left(0\right)-\frac{\ell}{3}x\right|\leq\varepsilon\left|x\right| $$
Which means $ f $ is differentiable at $ 0 $.
