Ant on a square There is a square of side $1 m$ and an ant has to cross diagonally. However, it chooses to walk along the boundary so the distance covered by it is $2m$ and not $\sqrt{2} m$. This it does in two moves. Now the ant decides to walk along the one side and stop at $\frac{1}{2}$ that is  $0.5 m $ and then go $0.5m$ up, $0.5m$ forward and another $0.5m$ upward to reach its destination. It does the same with $0.25 m$ ($\frac{1}{4})$etc. Assume now that the ant divides the segment $n$ times, that is, $\frac{1}{n}$ and covers the same distance, by $2n$ moves. As $n$ tends to infinity, the path of the ant will be equal to the straight line path that measures $\sqrt{2}m$. But then, if we consider this to be a normal limit problem as $n$ tends to infinity, the answer is always $2$. Where is the contradiction? I have tried drawing the paths and checking them for large values of $n$ but I still couldn't spot the contradiction. 
 A: We learn from this that the length of paths (or even their rectifiability) is not compatible with taking limits.
A: Basically, there is no contradiction. Let's see what you have done. You have constructed a line in space, $\Gamma$. It has a simple parametrisation of $\gamma:t\mapsto(t,t)$ for $t\in[0,m]$.
You then construct a sequence of paths $\Gamma_n$ which also have some parametrisation. For example, $\Gamma_1$ has parametrisation $\gamma_1:t\mapsto(2t,0)$ for $t\in[0,m/2]$ and $\gamma_1:t\mapsto(m, 2t)$ for $t\in[m/2,m]$. The other parametrisations are increasingly hard to write, but you basically just cut the two orthogonal lines and keep their parameter intact.
Now, it can be shown that in fact even the parametrisations $\gamma_i$ converge towards $\gamma$ and in fact do so uniformly. Your example therefore shows that even if $\Gamma$ is the limit of a family of functions, its length may not be.
So what is the problem? Why is the length not converging? In my mind, the most natural way of thinking is that the derivatives do not converge. A length of a line parametrised by $\gamma$ is defined as the integral of $|\dot\gamma|$ over the parametrising interval. If you have $\dot \gamma_n\rightarrow \dot\gamma$, then the integrals will converge. If you do not (and in your case you don't), there is nothing forcing the lengths to converge.
