Consider a function $\vec{f}:\Bbb R^2 \rightarrow \Bbb R^2$ with Frechet derivative $Df(\vec{a})$ at some $\vec{a} = (a_x, a_y) \in \Bbb R^2$, and infinitesimal rectangle $R(\delta x, \delta y) = [a_x, a_x + \delta x] \times [a_y, a_y + \delta y]$. Under transformation $\vec{f}$, this infinitesimal rectangle is mapped to an infinitesimal parallelogram and its area is scaled by the Jacobian $J = \lvert \det(Df(\vec{a})) \rvert$.
Formally, for $R(\Delta x, \Delta y) = [a_x, a_x + \Delta x] \times [a_y, a_y + \Delta y]$, $$ \lim_{\Delta \vec{x} \to \vec{0}} \frac{area(\vec{f}(R(\Delta \vec{x})))}{area(R(\Delta x, \Delta y))} = \lvert \det(D\vec{f}(a))) \rvert$$
Is it possible to formally prove this using the definition of the Frechet derivative? If not, how could it be done?
EDIT: The definition of $area(\vec{f}(R(\Delta \vec{x})))$ is the value such that for any $\epsilon > 0$ there exists partition $P$ of some rectangle enclosing $S = \vec{f}(R(\Delta \vec{x}))$ such that upper and lower Riemann sums $U_P \vec{f} \chi_S - L_P \vec{f} \chi_S < \epsilon$.
Thanks in advance for your help.
If $R$ is a rectangle, we can write the numerator $\mathrm{area}\ f(R)$ using an appropriate change of variables theorem (say for functions $f$ which are injective, $C^1$, with nonvanishing Jacobian $J_f$) with $y = f(x)$, $$ \mathrm{area}\ f(R) = \int_{f(R)}1\ dy = \int_{R} J_f(x) \ dx, $$ Now we see that dividing by the area of $R$, the fraction $\frac{\mathrm{area}\ f(R)}{\mathrm{area}\ R}$ is the integral average of $J_f(x)$ on the small rectangle $R$: $$ \frac{\mathrm{area}\ f(R)}{\mathrm{area}\ R} = \frac{1}{\mathrm{area}\ R}\int_R J_f(x)\ dx. $$ If $f$ is say $C^1$, then $J_f$ is continuous, so for any $x_0$ $$ \left|\frac{1}{\mathrm{area}\ R}\int_R J_f(x)\ dx - J_f(x_0)\right| = \left|\frac{1}{\mathrm{area}\ R}\int_R \big(J_f(x)-J_f(x_0)\big)\ dx\right|\leq \frac{1}{\mathrm{area}\ R}\int_R |J_f(x)-J_f(x_0)|\ dx. $$ Provided $x_0\in R$, and the diameter of $R$ is sufficiently small, this latter expression can be made arbitrarily small by continuity of $J_f$ at $x_0$.
The conclusion is if $f$ is sufficiently nice (say injective, $C^1$, with nonvanishing Jacobian $J_f$), then for each $x$ we have $$ \lim_{\mathrm{diam}\ R\to 0,x\in R}\frac{\mathrm{area}\ f(R)}{\mathrm{area}\ R} = J_f(x). $$