Why are these two definitions of differentiability identical? Recently, I have learned the following as the  definition of multivariable differentiability. Assume that one can express $f(x, y)$ in the following form:
$$f\left(x, y\right) = f\left(x_0, y_0\right) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \xi \left(x, y\right) $$
Where $\xi$ can be viewed as the "error" function between the surface and the tangent plane. 
Then for $f$ to be differentiable at $\left(x_0, y_0\right)$, the following must hold true for all $\epsilon > 0$:
For any $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $r < \delta$, $\xi < \epsilon r$
where $r = \sqrt{\left(x - x_0 \right)^2 + \left(y - y_0\right)^2}$.
In more intuitive terms, as we approach the point, $\xi$ needs to "disappear" faster than any linear multiple of $r$. If $\xi$ is linear in $r$, then we can say that $f$ is not differentiable (sort of analogous to a cusp in single variable calculus). This can also be stated as (I think):
$$\lim_{r \rightarrow 0} \frac{\xi(x, y)}{r} = 0$$
Intuitively, this makes sense to be, because when we get "infinitesimally" close to the point in question, we can utilize the tangent plane to obtain the differential expression:
$$\mathrm{d}f = \frac{\partial f}{\partial x} \mathrm{d}x + \frac{\partial f}{\partial y} \mathrm{d}y$$
But in the textbook I am using, differentiability is defined as follows:
$f(x, y)$ is differentiable at $(x_0, y_0)$ if we can express $\Delta f$ in the following manner:
$$\Delta f = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y$$
where as $\Delta x \rightarrow 0$ and $\Delta y \rightarrow 0$, $\epsilon_1 \rightarrow 0$ and $\epsilon_2 \rightarrow 0$.
You can view $\epsilon_1 \Delta x + \epsilon_2 \Delta y$ as the $\xi$ function mentioned in the first definition of differentiability. But I don't see how the two definitions are identical. It seems that in both of them, the "error" from the tangent plane is required to approach $0$ "quickly enough," but I don't quite see how the two are identical.
 A: Actually I don't think that was a good hint after all. I'll keep it below for laughs.
Let's simplify: I'll replace $\Delta x, \Delta y$ by $x,y$ and think of $(x,y) \to (0,0).$ Let $r=\sqrt {x^2+y^2}.$ The symbol $\epsilon$ will denote a function $\epsilon(x,y)$ such that $\lim\limits_{(x,y)\to (0,0)}\epsilon(x,y)=0.$
Suppose we have $\epsilon_1x + \epsilon_2y.$ Since $x=r\cos t, y = r\sin t,$ we have
$$\epsilon_1x + \epsilon_2y = [\epsilon_1\cos t  + \epsilon_2\sin t]r.$$
Clearly the function in brackets $\to 0,$ so we have written the left side as $\epsilon r$ as desired.
Now suppose we have $\epsilon r.$ Define $\epsilon_1 = \epsilon (r/x)$ if $|y|<|x|,\epsilon_1 = 0,|y|\ge |x|.$ Similarly, $\epsilon_2 = \epsilon (r/y)$ if $|y|\ge |x|,$ $\epsilon_2 = 0$ if $|y|<|x|.$ Check that
$$\epsilon r =  \epsilon_1x + \epsilon_2y.$$
We need to verify that $\epsilon_1,\epsilon_2$ do the job. But this follows because $r/x$ is bounded for $|y|<|x|$ and $r/y$ is bounded for $|y|\ge |x|.$

Earlier answer: Hint: If $a,b\in \mathbb {R},$ then $\sqrt {a^2+b^2} \le |a|+|b| \le \sqrt 2\sqrt {a^2+b^2}.$
