$\liminf_{\epsilon\to 0} \frac{f(\epsilon)}{\epsilon}$ finite implies $\lim_{\epsilon \to 0} f(\epsilon)=0$ Let $f:\mathbb{R}\to\mathbb{R}$ be a non negative real function. 
Question: Is it true that if $\liminf_{\epsilon\to 0} \frac{f(\epsilon)}{\epsilon}$ exists finite than necessarily $\lim_{\epsilon \to 0} f(\epsilon)=0$ ?

Note that for any $x_n \to 0$, 
$$f(x) = \begin{cases} 0 & \text{if } x = x_n \\ \text{anything }\ge 0 &\text{otherwise}.\end{cases}$$
satisfies 
$$\liminf_{\epsilon \to 0} \frac{f(\epsilon)}{\epsilon}  =0.$$
