Let f be a map from $\mathbb{R}^2$ to $\mathbb{R}^2$ be defined as $f(x,y)=(x^2,y^2)$ let T from $\mathbb{R}^2$ to $\mathbb{R}^2$ be the map $T(x,y)=(2x,4y)$ then to show that f is differentiable at $(1,2)$ with derivative T, the following change of variables is made:
$$\lim_{(x,y) \rightarrow (1,2)}\frac{\|f(x,y)-f(1,2)-T((x,y)-(1,2))\|}{\|(x,y)-(1,2)\|}$$
Set $(x,y)=(1,2)+(a,b)$ we have the above expression is equal to
$$\lim_{(a,b) \rightarrow (0,0)}\frac{\|f(1+a,2+b)-f(1,2)-T((a,b)\|}{\|(a,b)\|}$$
The question I have is – why are we allowed to make the substitution? How can we be sure that the two expressions are equal? Can this be generalized? What lemma or Theorem is involved with regards to this substitution and equality? I'm trying to understand this at a formal level, rather than on an intuitive one.
Asserting that$$\lim_{(x,y)\to(1,2)}\frac{\left\lVert f(x,y)-f(1,2)-T\bigl((x,y)-(1,2)\bigr)\right\rVert}{\bigl\lVert(x,y)-(1,2)\bigr\rVert}=0$$is equivalent to$$(\forall\varepsilon>0)(\exists\delta>0):\sqrt{(x-1)^2+(y-2)^2}<\delta\implies\frac{\left\lVert f(x,y)-f(1,2)-T\bigl((x,y)-(1,2)\bigr)\right\rVert}{\bigl\lVert(x,y)-(1,2)\bigr\rVert}<\varepsilon.\tag1$$And asserting that$$\lim_{(a,b)\to(0,0)}\frac{\bigl\lVert f(1+a,2+b)-f(1,2)-T(a,b)\bigr\rvert}{\bigl\lVert(a,b)\bigr\rVert}=0$$is equivalent to$$(\forall\varepsilon>0)(\exists\delta>0):\sqrt{a^2+b^2}<\delta\implies\frac{\left\lVert f(1+a,2+b)-f(1,2)-T(a,b)\right\rVert}{\bigl\lVert(a,b)\bigr\rVert}<\varepsilon.\tag2$$It is clear that $(1)$ and $(2)$ are equivalent: for each $\varepsilon>0$, any $\delta$ that works for $(1)$ also works for $(2)$ and vice-versa.