# 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.

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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 $h<0$.

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

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I am sorry, it should have been a strict inequality. I realize why should it be. Thanks for pointing that out! – user43901 Jan 9 '13 at 22:01
This is allright. While we are at it, I find funny that you accepted the only answer I do not understand. Surely you can tell me why this answer asserts that $\lim f(h)/h=\lim |f(h)|/h$. – Did Jan 10 '13 at 0:03
I did not accept your answer since it was not a complete answer. As far as the point you mention, I wonder why you think that is wrong given the problem statement. Surely you can clarify that. – user43901 Jan 10 '13 at 19:02
My comment did not ask why you did not choose my answer (which is not incomplete, thank you) but why you accepted the answer you accepted. Since you accepted it, of course you understood why $\lim f(h)/h=\lim |f(h)|/h$ (to me, this is not obvious since $\lim \varphi(h)=\lim|\varphi(h)|$ does not hold in general). Surely you can clarify that. – Did Jan 10 '13 at 20:25
I think you are taking this too personally. Perhaps "incomplete" was not the right choice of word; I would have rather preferred a more detailed answer. You just wrote the expression and then wrote infinity (which confused me even more). In this case, I do think that the limits equal given the problem statement. – user43901 Jan 11 '13 at 0:14

$$\left|\frac{f(h)}{h}\right|\geq |h|^{\beta-1}.$$ 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.

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