# Prove $f$ is not differentiable at $(0,0)$

Mathematics Asked by Kelan on February 27, 2021

For
$$f(x,y)=begin{cases} frac{x|y|}{sqrt{x^2+y^2}} & text{ for }(x,y)neq (0,0)\ 0 & text{ for } (x,y)=(0,0) end{cases}$$

I’m trying to prove $f$ is not differentiable at $(0,0)$. I showed if $f$ is differentiable at $(0,0),$ then $A=Df_{(0,0)}=0.$ But I don’t know how this lead to a contradiction. Anyone has ideas?

If you approach with the paths $(t,0)$ and $(0,t)$, you will conclude that the partial derivatives are 0. Thus in order to show that it is not differentiable at $(0,0)$ it suffices to show a path that leads to a different derivative.

One path you may consider is $(t,t)$: For $t neq 0$, $f(t,t)=frac{t |{t}|}{sqrt{2t^2}} =frac{t {|t|}}{sqrt{2} {|t|}} =frac{t}{sqrt{2}}$ which leads a derivative of $frac{1}{sqrt{2}} neq 0$.

Answered by Carlos Pinzón on February 27, 2021

I'll show that $f$ isn't continuous. Consider the line $y=x$. Now,

$lim_{(x,x)to (0^+,0^+)} f(x)=1/sqrt{2}neq 0$.

$lim_{(x,x)to (0^-,0^-)} f(x)=-1/sqrt{2}neq 0$.

Because I found two directions with distinct limits, the function isn't continuous (and more, we can't redefine $f$ because the limit don't exist)

Answered by Martín Vacas Vignolo on February 27, 2021