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

For every $\{t_k\}\subset \mathbb{R}$, there exists a real number $\alpha$, depending on the sequence $\{t_k\}$, such that, if we set $u_k(x)={u(t_k x) \over t_k}$, we have that $u_k(x) \to \alpha x$ 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. Does the answer change if the uniform convergence holds for a subsequence of $u_k$?

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.