Computing diagonal Length of a Square While studying rectification of curves, I considered a curve and to measure its length in a different fashion, and arrived at a problem. I would like to clarify the confusion in my understanding. 
Consider a unit square in plane with vertices (0,0), (0,1),(1,0), (1,1). The diagonal joining (1,0) and (0,1) has length $\sqrt{2}$, well known. Suppose I approach theis diagonal in the following way: first by the path $P_1: (1,0)-(1/2,0)-(1/2,1/2)-(0,1/2)-(1,0)$. This is like tow "L"s, with top end of one joined to bottom end of other- forming stairs. The length of this path is $2$. Next, we form path $P_2$ with four "L"'s, in a nice way to form stairs. Again the length of this path is $2$. 
We see that the sequence $\{P_n\}$ of paths, which are piecewise differentiable functions (?), converges to the diagonal $(1,0)-(0,1)$. But the length of each path is $2$, but we cant conclude that the diagonal should have length $2$. Why such a contradiction arises?
 A: the short answer is that length is not a continuous operation. Even a a sequence of of paths converges uniformly, the limit of the lengths of the curves need not converge to the length of the limiting curve. The most you can say is that length is lower semi-continuous, meaning length of the limiting curve is a lower bounder for the limit of the lengths of the curves. Your construction illustrates just that. 
The moral is that length is a tricky thing. Very tricky in fact. It is common to define the length of the graph of a function $f(t)$ by the integral $\int_a^b(1+f'(t)^2)^{1/2}$, arriving at this formula by an approximation of the curve by straight lines and using Pythagoras. In fact, we take this as the definition of the length with the derivation as motivation, not an actual derivation (in a sense). The example you gave should at the very least indicate that such a definition should be looked at with some scrutiny and not accepted too easily. If you do accept it, then the subtlety of lengths is easily seen to be related to a similar subtlety of derivatives. It is well-known that if a sequence of functions converges uniformly, then the limits of the derivatives need not converge to the derivative of the limit. The reason is obvious, the derivative is extremely sensitive to tiny perturbations, while uniform convergence is not. 
This is exactly what happens with the length. For a more extreme example than the one you gave, consider $\frac {1}{n}\sin(1/x^n)$ between $0$ and $1$. This sequence converges to the constantly $0$ function, but the oscillations become more and more rapid. As a result the graphs become longer and longer, and actually approaches infinity. Of course the length of the limiting function is just $1$.
A: You cannot compute the length by taking any sequence of paths that converge uniformly.
To illustrate, suppose I have a straight line 1km long. Instead of walking straight, I walk at an angle $\theta$ to the direction of forward motion. After I travel some distance $\delta>0$, I walk at an angle $-\theta$ for another $\delta$ then repeat until I reach the end. It should be clear that I will cover a distance $\approx { 1 \over \cos \theta}$km. Hence I can choose $\theta$ to make the distance I cover as large as I want, and have $\delta$ be small enough so I remain arbitrarily close to the straight line.
The issue is that the arc length function $l : C[0,1] \to [0,\infty)$ is not continuous (with norm $\|f\|_\infty = \max_{t \in [0,1]} \|f(t)\|$).
If we choose a more restrictive space and topology we can improve on this situation. One possibility is to choose the space of continuously differentiable functions $C^1[0,1]$ with the norm $\|f\|_* = \|f\|_\infty+\|f'\|_\infty$. Since we have $l(f) = \int_0^1 \|f'(t)\| dt$, then we see that $l$ is continuous from the estimate
$|l(f)-l(g)| \le \int_0^1 | \|f'(t)\| - \|g'(t)\| | dt \le \int_0^1 \|f'(t) - g'(t)\| dt \le \|f-g\|_*$.
