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.