In proving the sufficient condition, Ahlfors claims that if $u(x,y)$ has a continuous, first order partials, then we can write $$u(x+h,y+k) - u(x,y) = \frac{\partial{u}}{\partial {x}}h +\frac{\partial{u}}{\partial {y}}k + \epsilon$$ where $\epsilon$ is little-o of $h +ik$. Now I am familiar with this definition for real functions of one variable (in fact I attempted a proof of it on a previous question), however I'm having difficulties seeing it given the limit definition of partials. Can anyone show a quick proof of this for 2 or n-variables.
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1You lost a $k$ in there. – GEdgar Jul 30 '18 at 20:56
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Continuous partial derivatives implies differentiability, can be found in many books of multivariable calculus, and has been asked and proved here before.
Differentiability implies that it can be written in such a form, with some linear function on $(h,k)$ in place of $\frac{\partial u}{\partial x}h+\frac{\partial u}{\partial y}k$.
$$\begin{align}u(x+h,y+k)-u(x,y)&=L_{(x,y)}(h,k)+\epsilon\\L_{(x,y)}(h,k)&=ah+bk\end{align}$$
Restricting the limit in the definition of differentiability to the $X$-axis and then to the $Y$-axis you get that the linear function has coefficients $a=\frac{\partial u}{\partial x}$ and $b=\frac{\partial u}{\partial y}$.

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