Differentiability of $f$. Given $0 < \beta < 1$ and $|f(x)|\geq |x|^ \beta $ and $f(0)=0$. Show that $f$ is not differentiable at $0$ This is what I have so far. I think it is the absolute value signs that are throwing me off. By the definition of derivatives,

$$
f'(0) =  \lim_{h \to 0} \frac {f(0 +h) - f(0)} {h}= \lim_{h \to 0} \frac {f(h)} {h}\\
$$

But now what? Since $f(0)=0$ and as $h$ tends to $0$, doesn't the the derivative read $0/0$? 
I know that I should be getting the left and the right hand limit to be different at zero. I am just failing to carry the rest of the proof through. Will appreciate any help.
The result is wrong if $\beta=1$ (try $f(x)=x$). If $\beta\lt1$, this follows from
$$
\left|\frac{f(h)}h\right|\geqslant\frac1{|h|^{1-\beta}}\to\infty.
$$
Since as you stated:
$$f'(0) = \lim_{h\rightarrow 0} \frac{f(h)}{h}=\lim_{h\rightarrow 0} \frac{|f(h)|}{h}\ge\lim_{h\rightarrow 0} \frac{|h|^\beta}{h}$$
If we approach from the right, $h&gt;0$. Assume $f(h)&gt;0, |\beta|&lt;1$:
$$f'(0)=\lim_{h\rightarrow 0} \frac{|f(h)|}{h}\ge\lim_{h\rightarrow 0} \frac{|h|^\beta}{h}=\lim_{h\rightarrow 0} h^{\beta-1}=\infty$$
If $f(h) &lt;0$, we get the same divergence but to $-\infty$. The same proof follows for $h&lt;0$.