[fiforpg]

Nice example. You can avoid the issue with these being non-functions (because of the vertical segments) by rotating the picture so that the (0,0)-(1,1) diagonal is horizontal. That way you are talking about a sequence of continuous functions F_n with slopes alternating between ±1 that approach the constant-zero function F_0=0.

The reason your intuition was confused was that you felt that there was a way in which these functions with ±1 slope converge to zero function (it is called uniform convergence), but were unsure about the way in which their derivatives f_n=(F_n)' (which keep flipping between ±1) converge to zero.

The resolution of the seeming difficulty is that there are different modes of convergence. The ±1 slope functions F_n converge uniformly to zero, but their derivatives f_n, which determine curve length, do not converge uniformly or even pointwise to any limit (uniform convergence is stronger — more restrictive — than pointwise). This is why it is reasonable that the F_n curve length stays 2 throughout the exercise.

Note: if the derivatives f_n were converging to f_0=(F_0)'=0 pointwise while staying bounded, then the curve lengths of F_n would converge to sqrt(2), the curve length of F_0. This is called "dominated convergence theorem" for integrals — point being that the curve length of F_n is an integral of sqrt(1+f_n^2). But there is no pointwise convergence, and no such implication.

Finally, you may be curious as to whether there is a sense in which f_n converge to f_0=0 — does all that flipping back and forth amount to some way of converging to zero? Turns out yes! It is called weak* (weak-star) convergence and it applies to f_n if you think of them as measures (or distributions) — that is, if instead of pointwise evaluation you characterize f_n by the way they act on test functions by integration: instead of computing f_n(x) you multiply f_n * phi and integrate. Phi has to be a continuous function. Under this notion of evaluation, f_n do converge to zero.

Edit: typos.