Let $u \in C^0([-1,1])$ such that $u(0)=0$. Suppose that $u$ satisfies the following property:


For every $\{t\_k\}\subset \mathbb{R}$ such that $t\_k \to 0$, there exist a real number $\alpha$ (depending on the sequence $\{t\_k\}$), and a subsequence $\{t\_{k\_n}\}$, such that, if we set $u\_k(x)={u(t\_k x) \over t\_k}$, we have that $u\_{k\_n}(x) \to \alpha x$ uniformly on $[-1,1]$.


My question is the following:
Is the function $u$ necessarily differentiable at the origin? Even if $\alpha$ depends on the sequence $\{t\_k\}$, I can't imagine how such a counterexample should behave.



Differentiability of the blow-up of a function
Set $f(x)=x\cos(\log(\log(1/|x|)))$ (and 0 at 0). 


Let $\epsilon>0$. Now if $0<t<\epsilon^{1/\epsilon}$, consider 
$\Delta\_t(x):=|f(tx)/t - x\cos(\log\log(1/t)|$.


On $[-\epsilon,\epsilon]$, this is bounded above by $2\epsilon$. 


If $\epsilon<|x|\le 1$, we have
\begin{align\*}
\Delta\_t(x)&=|x[\cos(\log\log(1/(t|x|))-\cos(\log\log(1/t))]|\\
&\le \log\log(1/t|x|)-\log\log 1/t\\
&= \log(\log(1/t)+\log(1/|x|))-\log\log(1/t)\\ &\le \log(1/|x|)/\log(1/t)<\epsilon,
\end{align\*}
where I used the concavity of $\log$ in the inequality before last. 


