```Date: Nov 18, 2013 6:45 PM
Author: Hetware
Subject: Proof that mixed partials commute.

In _Calculus and Analytic Geometry_ 2nd ed.(1953), Thomas provides:Theorem. If the function w=f(x,y) together with the partial derivatives f_x, f_y, f_xy and f_yx are continuous, then f_xy = f_yx.Both Thomas and Anton (1980) provide rather long-winded proofs of this theorem.  These proofs involved geometric arguments, auxiliary functions, the mean-value theorem, epsilon error variables, a proliferation of symbols, and a generous helping of obscurity.Starting from the definition of partial differentiation, and using the rules of limits, along with a modest amount of basic algebra, I came up with this:f_x(x,y) = Limit[[f(x+Dx,y)-f(x,y)]/Dx, Dx->0]f_yx(x,y) = Limit[[f_x(x,y+Dy)-f_x(x,y)]/Dy, Dy->0]  = Limit[[[f(x+Dx,y+Dy)-f(x,y+Dy)]-[f(x+Dx,y)-f(x,y)]]/DyDx, {Dy->0, Dx->0}]f_xy(x,y) = Limit[[[f(x+Dx,y+Dy)-f(x+Dx,y)]-[f(x,y+Dy)-f(x,y)]]/DxDy, {Dx->0, Dy->0}]The only caveat is that the rules for limits, such as /the product of limits is equal to the limit of the products/, are stated in terms of a single variable. For example:Limit[F(t) G(t), t->c] = Limit[F(t), t->c] Limit[G(t), t->c]Whereas I am assumingLimit[F(x) G(y), {x->c, y->c}] = Limit[F(x), x->c] Limit[F(y), y->c].I argue as follows. The statement that x->c as y->c can be formalized by treating x and y as functions of t such thatLimit[x(t), t->c] = Limit[y(t), t->c] = c|F(x)-F(c)| < epsilon_F ==>  |x-c| < delta_F exists|G(y)-G(c)| < epsilon_G ==>  |y-c| < delta_G exists|x(t)-c| < epsilon_x ==>  |t-c| < delta_x exists|y(t)-c| < epsilon_y ==>  |t-c| < delta_y existsNow epsilon_F ==> delta_F can be used as delta_F = epsilon_x which implies delta_x > |t-c| exists. SoLimit[F(x(t)), t->c] = Limit[F(x), x->c], etc.It follows thatLimit[F(x(t)) G(y(t)), t->c]= Limit[F(x(t)), t->c] Limit[G(y(t)), t->c]= Limit[F(x), x->c] Limit[G(y), y->c]Am I making sense here?  I feel as though I am trying to prove the obvious, but it is not obvious how to prove it.
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