The usual definition of differentiability for a multivariable function says that $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at $(x,y)$ if there is a linear map $d_{(x,y)} f : \mathbb{R}^2\to \mathbb{R}$ such that
$$ \lim_{(h,k)\to 0} \frac{f(x+h,y+k) - f(x,y) - d_{(x,y)}f(h,k)}{\Vert(h,k)\Vert} = 0. $$
This is equivalent to asking that
$$ f(x+h,y+k) = f(x,y) + d_{(x,y)}f(h,k) + \epsilon(h,k) \Vert(h,k)\Vert $$
where $\epsilon(h,k)\to 0$ as $(h,k)\to 0$.
However, the calculus book that I'm teaching out of this semester defines differentiability differently, asking that
$$ f(x+h,y+k) = f(x,y) + d_{(x,y)}f(h,k) + \epsilon_1(h,k) h + \epsilon_2(h,k) k $$
where $\epsilon_1(h,k)\to 0$ and $\epsilon_2(h,k)\to 0$ as $(h,k)\to 0$. Why is this an equivalent definition? I see why the second implies the first: the obvious definition $\epsilon = \frac{\epsilon_1 h + \epsilon_2 k}{\Vert(h,k)\Vert}$ works because $\frac{|h|}{\Vert(h,k)\Vert} \le 1$ and $\frac{|k|}{\Vert(h,k)\Vert} \le 1$. But I don't see the converse: given $\epsilon(h,k)$, how do I define $\epsilon_1$ and $\epsilon_2$?
