Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2?
I thought that if there exists a limit on both sides then it would be differentiable?
[Math] Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2
calculusderivativeslimits
Related Solutions
There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
Write $f(x)=x$ if $x\geq0$ and $f(x)=2x$ if $x<0$.
To make the second conclusion, you would need $f'$ to exist throught the open interval $(-1,1)$, which is obviously not true here.
Best Answer
Here's the graph of $f:=\sqrt{\vert x-2 \vert}$ where $-5 \leq x \leq 5$:
Here $f$ is a continuous curve but there is a "corner" at $x=2$, so $f$ is not differentiable there.
For the formal argument, the limit $$\lim_{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim_{h \rightarrow 0} \frac{\sqrt{\vert 2+h-2 \vert} \;-\sqrt{\vert 2-2 \vert} }{h}=\lim_{h \rightarrow 0} \frac{\sqrt{\vert h \vert}}{h} $$ does't exist!