Arc length of a sequence of semicircles. Let $\gamma_1$ be the semicircle above the x axis joining $-1$ and $1$.
Now divide this interval into $[-1,0]$ and $[0,1]$, and trace a semicircle joining $-1,0$ above the $x$-axis and another one joining $0,1$ below the $x$-axis, call this curve $\gamma_2$
Keep subdividing the interval into $2^n$ subintervals, proceding as above, and call the resulting curve $\gamma_n$.
Now, from looking at their graphs, I can say that the arc length of all these curves is $\pi$, but the arc length of the limit, $\gamma$ (which I think should be a straight line connecting $-1$ and $1$) is $2$.
Why does this happen? Shouldn't the arc length of the limit be $\pi$ as well? Does $\gamma_n\to \gamma$ uniformly?
One of my professors said that this is because
$$
\operatorname{Arc}(\gamma_i)=\int_D\sqrt{1+\gamma_i'^2}dx
$$
And that $\gamma_i'$ becomes pretty crazy when $i$ is large (as it's $\pm \infty$ on the edges of the circles).

$\lim \limits_{n\to\infty}{2^n*Pi/2^n}$
This obviously is $Pi$. The circles may seem very small, but can't get $0$ (otherwise, the two points at $-1$ and $1$ wouldn't be connected). Consider their radius' as infinitesimal, so the curve length always will stay $Pi$ and will never get $2$.
Also, the the length of $2$ would mean that the curve is a straight line and accordingly that it contains all points with $y=0$ and $-1\le x\le1$.
There are simply $2^n+1$ points that touch the x-axis (so the x-coordinates are rational), infinitely many, but not all. This might sound contradictory, however, Cantor has proven that there are more real than rational numbers (Cantor's Diagonal Argument).
