How to determine if $\ f(x)=|x-2| $ is differentiable at 2 I need to determine if $\ f(x)=|x-2| $ is differentiable at 2.
I was thinking I could use the definition of a derivative $\ \frac{(f(x+h)-f(x))}{h} $ but am kind of at a loss.
 A: Hint:
$$\lim_{h\to0}\frac{f(2+h)-f(2)}h=\lim_{h\to0}\frac{|h|}h=\begin{cases}+1,&h\to0^+\\-1,&h\to0^-\end{cases}$$
A: Your idea of considering the definition of derivative is a good one.
Indeed, not that for $h>0$ we have $|h|=h$ and for $h<0$ it holds $|h|=-h$. It follows that
$$1=\lim_{h\to 0^+}\frac{h}{h}=\lim_{h\to 0^+}\frac{|h|}{h}=\lim_{h\to 0^+}\frac{f(2+h)-f(2)}{h} $$ and $$-1=\lim_{h\to 0^-}\frac{-h}{h}=\lim_{h\to 0^-}\frac{|h|}{h}=\lim_{h\to 0^-}\frac{f(2+h)-f(2)}{h}$$
Hence, the limit 
$$\lim_{h\to 0}\frac{f(2+h)-f(2)}{h} $$
does not exists and thus the function is not differentiable at $2$.
A: By definition, $f(x) = x-2$ when $x \ge 2$ and $f(x) = - (x-2) = -x+2$ when $x < 2$.
Then $f'(x) = 1$ when $x > 2$ and $f'(x) = -1$ when $x < 2$.
So, what is the derivative at $x=2$? It isn't defined since the slope isn't the same approaching from both sides.
That is, the limit
$$f'(2) := \lim_{h\to0} \frac{f(2+h) - f(2)}{h}$$
does not exist since the left and right hand limits do not match.
A: The function is $f(x)=x-2$ for $x\ge 2$ and $f(x)=2-x$ for $x<2$, so at $x=2$ we have a left derivative that is $f'_-(2)=-1$ and a right derivative $f'_+(2)=1$. Since they are different the function is not differentiable at $x=2$.
