The given function is $f(x,y) = \lvert x^2 - y^2 \rvert$
The question is to verify that $f_{xy} = f_{yx}$ at $(0,0)$
The solution has one part that I'm unable to comprehend
It starts with
$$ f_{xy}(0,0) = \lim \limits_{h \to 0} \frac{f_y(h,0) - f_y(0,0)}{h}$$
Where we have
$$ f_y(0,0) = \lim \limits_{k \to 0} \frac{f(0,k) - f(0,0)}{k} = \lim \limits_{k \to 0} \frac{\lvert k^2 \rvert}{k} = 0$$
And $$ f_y(h,0) = \lim \limits_{k \to 0} \frac{f(h,k) - f(h,0)}{k} $$
$$ = \lim \limits_{k \to 0} \frac{\lvert h^2 - k^2 \rvert - \lvert h^2 \rvert}{k} $$
Here they have provided the condition that if $ h^2 > k^2 $
Then
$$ = \lim \limits_{k \to 0} \frac{(h^2 - k^2) - h^2}{k} = 0$$
And proceeded further.
My doubt is precisely in this step. How can they just take only the value of $ h^2 > k^2$ and move on? What about if $ k^2 > h^2$? If the latter is true then won't this imply that the partial differential $f_{xy}(0,0)$ doesn't exist?
They have also similarly assumed if $h^2 < k^2$ to prove existence of $f_{yx}(0,0)$.
According to my understanding the values of the variables of $h,k$ (taken for the limits) vary independently of each other. So why have they taken these assumptions?
Ok so as Andrei mentioned in the comment , while one of the values of $h$ or $k$ is tending the other non-tending value however small it may be, will always end up greater than the tending value, as the tending value tends to zero.