Prove that $f(x,y)=1- \sqrt{|xy|}$ is not differentiable at $(0,0)$

Question

Let $f(x,y)=1- \sqrt{|xy|}$. Prove that $f$ is not differentiable at $(0,0)$.

I don't really know how to tackle this problem. Normally for this kind of problem I would find the candidate to be differential, this is, $L(x,y) = \frac{\partial f}{\partial x}(0,0)x + \frac{\partial f}{\partial y}(0,0)y$ and I would try to find out if $\displaystyle\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L(x,y)}{ \| (x,y)\|}=0$. But I don't know how to start being $f$ differentiable.

Answer 1

Take the directional derivative of $f$ along the direction $(1,1)$. For all points of the form $(t,t)$, we have $$\frac{f(t,t) - f(0,0)}{t} = \frac{1 - \sqrt{\vert t^2 \vert} - 1}{t} = \frac{- \vert t \vert}{t}$$

The limit as $t \rightarrow 0$ does not exist because: Take $t = + \epsilon$ where $\epsilon >0$, the limit is $-1$ as $\epsilon \rightarrow 0$. Now, take $t = -\epsilon$ where $\epsilon >0$ and let $\epsilon \rightarrow 0$, the limit will be $1$. So two different limits while you're going to zero. Hence, limit doesn't exist.