prove that $f(x,y)=|xy|$ is differentiable at (0,0) using $\epsilon_1,\epsilon_2$ definition I would like to prove the differentiability of $f(x,y)=|xy|$ at $(0,0)$ using the $\epsilon_1,\epsilon_2$ definition ( from thomas calculus)

My Working:
$f_x (0,0)=\lim_{h\to 0} \frac{f(h,0) - f(0,0)}{h} = 0$
similarly $f_y (0,0)=0$
$\Delta z= |\Delta x||\Delta y|$
Hence by definition we must represent $\Delta z$ as $\epsilon_1 \Delta x + \epsilon_2 \Delta y $
$|\Delta x||\Delta y|=\epsilon_1 \Delta x + \epsilon_2 \Delta y $
but I am unable to find any such $\epsilon_1,\epsilon_2$ and moreover they must tend to 0 as $\Delta x, \Delta y$ tend to 0.
I know you can prove it using the total derivative definition but what's going wrong here?
 A: I think we can just take
$$
\epsilon_1
=
\frac{|\Delta x \Delta y|}{2\Delta x}\\
\epsilon_2
=
\frac{|\Delta x \Delta y|}{2\Delta y}
$$
$\newcommand{\sgn}{\textrm{sgn}}$
Recalling that $|a| = \sgn(a)a$, for all nonzero $\Delta x $ and $ \Delta y$,
$$
\epsilon_1 = \frac{1}{2}\sgn(\Delta x)|\Delta y|
$$
and $|\epsilon_1| = |\Delta y|/2$. Therefore,
$$
\lim_{\Delta x \to 0}\left(\lim_{\Delta y \to 0} \epsilon_1\right) = 0
$$
The process similarly applies to $\epsilon_2$.
A: I would recommend computing partial derivatives and then checking that the limit definition you have works out. In particular,
$$f_x(0,0) = \lim_{\Delta x\to 0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x} = \lim_{\Delta x\to 0} \frac 0{\Delta x} = 0,$$
and similarly for $f_y$. So you need to look at
$$\Delta z = |\Delta x\Delta y| - 0\Delta x - 0\Delta y = |\Delta x\Delta y|.$$
Now this can be split up in zillions of different ways as $\epsilon_1\Delta x+\epsilon_2\Delta y$. For example, take
$\epsilon_1 = \frac12\eta_1|\Delta y|$ and $\epsilon_2 = \frac12\eta_2|\Delta x|$, where $\eta_1 = \dfrac{|\Delta x|}{\Delta x}$ whenever $\Delta x\ne 0$ (and say $0$ when $\Delta x = 0$), and similarly for $\eta_2$. This is actually quite an awkward definition to work with in this case because of the absolute values.
For your reference, a more common definition, which is easier to work with, is that we should have
$$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{\Delta z - f_x(x_0,y_0)\Delta x - f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0.$$
In our case, we only have to check that
$$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{|\Delta x\Delta y|}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0.$$
Can you do this?
