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.

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.