# For $f(x)=\sqrt{|x-1|}$, show that the limit $\lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ not exist.

Using deltas and epsilons...

$f(x)=\sqrt{|x-1|}$. Show that $\lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ does not exist.

Well, I tried to get the fraction into x-a form, and I got to $\frac{|h|-2}{h\sqrt{|h|}+\sqrt{2}}$. I was going to set it to less than $\epsilon$, and show that $|f(x)-L|$ is greater/equal to zero.

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I've made some latex edits to your question. Please review and ensure that it's correct. – Calvin Lin Jun 2 '13 at 22:35

Note that $$\frac{f(1+h)-f(1)}{h}=\frac{\sqrt{ |h|}}{h}$$

since $\sqrt{|1+h-1|}=\sqrt{|h|}$ and $\sqrt{|1-1|}=\sqrt 0 =0$

What is this for $h>0$? And for $h<0$? You should get $+\infty$ and $-\infty$. Have you tried plotting this function near $x=1$ and see what is going on?

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I don't understand how I should put that according to the epsilon-delta definition. Do I have to prove that there doesn't exist a delta for any epsilon I choose? – Rubbles Jun 3 '13 at 17:11