$\lim_{\epsilon \to 0^+}(\phi(0)-\phi(\epsilon))\ln\epsilon=0$ Let $\phi:\mathbb{R}\to \mathbb{R}$ be a continuous differentiable function. Is it true that 

$$\lim_{\epsilon \to 0^+}(\phi(0)-\phi(\epsilon))\ln\epsilon=0?$$

I think this can be proved through an infinitesimal comparison, but I have no idea how to start with.
Any hint would be really appreciated.

An idea:
$$\lim_{\epsilon\to0}\frac{\phi(\epsilon)-\phi(0)}\epsilon=\phi'(0)$$
so
$$\lim_{\epsilon\to0^+}\left(\phi(0)-\phi(\epsilon)\right)\log\epsilon=\lim_{\epsilon\to0^+}\frac{\phi(0)-\phi(\epsilon)}\epsilon\epsilon\log\epsilon=-\phi'(0)\lim_{\epsilon\to0^+}\epsilon\,\log\epsilon=0$$
The one-before-the last step is due to the fact that both limits exist and are finite.
