Equivalence of Differentiability of Real Functions of Several Variables with Lame MTH254 Definition

(Necessary exercise) Let $f : ℝ^2→ℝ$, fix $α∈ℝ^2$, and define $Δf=f(α+(Δx,Δy))-f(α)$ given $Δx,Δy∈ℝ$. Show that the condition (1) The partial derivatives $f_x=\dfrac{\partial f}{\partial x},f_y=\dfrac{\partial f}{\partial y}$ exist at $α$ and there exists $ε_1,ε_2 : ℝ^2→ℝ$ such that $$Δf=f_x(α)Δx+f_y(α)Δy+ε_1Δx+ε_2Δy$$ $$ε_1,ε_2→0~~~~~~~~~\text{as}~~Δx,Δy→0$$ (implicitly $ε_i=ε_i(Δx,Δy)$) is equivalent to the condition (2) There exists a linear transformation $A : ℝ^2→ℝ$ such that when $h=(Δx,Δy)$, we see $$\lim_{h→0} \dfrac{|Δf-Ah|}{|h|} = 0$$ Proof: We shall use the condition—equivalent to (2)—of there existing a linear transformation $A: ℝ^2→ℝ$ and an error term $r : ℝ^2→ℝ$ such that $$Δf=Ah+r(h)$$ $$\lim_{h→0} \dfrac{|r(h)|}{|h|}→0$$ in the proof that follows. (1)$⇒$(2) Define $A(Δx,Δy)=f_x(α)Δx+f_y(α)Δy$, and $r(Δx,Δy)=ε_1Δx+ε_2Δy$. Then clearly $Δf=Ah+r(h)$, and also $$\dfrac{|r(h)|}{|h|}=|ε_1\dfrac{Δx}{|h|}+ε_2\dfrac{Δy}{|h|}|≤|ε_1|+|ε_2|→0$$ (2)$⇒$(1) Let $α=(x_0,y_0)$. Observing the real functions $x↦(x,y_0)$ and $y↦(x_0,y)$, we see by application of (2) that the appropriate partial derivatives exist, and that such a linear transformation must in fact be $A(Δx,Δy)=f_x(α)Δx+f_y(α)Δy$. Therefore define $$ε_1 = \left\{ \begin{array} \{ r(h)/Δx & Δx≠0 \\ 0 & Δx=0,Δy≠0 \\ 0 & Δx,Δy=0 \end{array} \right.~~~~~ε_2 = \left\{ \begin{array} \{ 0 & Δx≠0 \\ r(h)/Δy & Δx=0,Δy≠0 \\ 0 & Δx,Δy=0 \end{array} \right.$$ then it is clear that $Δf=f_x(α)Δx+f_y(α)Δy+ε_1Δx+ε_2Δy$, and observing $$|\dfrac{r(h)}{Δx}|≤|\dfrac{r(h)}{Δx}·\dfrac{Δx}{|h|}|=\dfrac{|r(h)|}{|h|}→0$$ and similarly $\dfrac{r(h)}{Δy}→0$ we see $ε_1,ε_2→0$ as $Δx,Δy→0$.$~\square$