Differentiability of $f$. Given $0 < \beta < 1$ and $|f(x)|\geq |x|^ \beta $ and $f(0)=0$. Show that $f$ is not differentiable at $0$

Question

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.

Answer 1

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, $h>0$. Assume $f(h)>0, |\beta|<1$: $$f'(0)=\lim_{h\rightarrow 0} \frac{|f(h)|}{h}\ge\lim_{h\rightarrow 0} \frac{|h|^\beta}{h}=\lim_{h\rightarrow 0} h^{\beta-1}=\infty$$ If $f(h) <0$, we get the same divergence but to $-\infty$. The same proof follows for $h<0$.

Answer 2

The result is wrong if $\beta=1$ (try $f(x)=x$). If $\beta\lt1$, this follows from $$ \left|\frac{f(h)}h\right|\geqslant\frac1{|h|^{1-\beta}}\to\infty. $$

Answer 3

$$ \left|\frac{f(h)}{h}\right|\geq |h|^{\beta-1}. $$ If $0\leq\beta< 1$, we have that $\beta-1$ is negative. Hence $|h|^{\beta-1}$ diverges as $h$ tends to zero and the limit $\lim_{h\to 0}f(h)/h$ does not exist.