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).
 A: The problem with this argument, and many other alike, is that the arc length of a curve is not a continuous function of a curve, which means that if sequence of curves $\gamma_i$ converges to a curve $\gamma$ (notion of convergence of functions like that can be made formal, but let me not do that here), then there is no guarantee the sequence of arclengths $l(\gamma_i)$ to converge to the arclength $l(\gamma)$. Indeed, if you look at the integrals $\int_0^1\sqrt{1-\gamma_i'(t)^2}dt$, then you might expect that the expression under the integral converges to $\sqrt{1-\gamma'(t)^2}$, which however isn't the case, because convergent sequence of differentiable functions (I am ignoring the "cusps" where the semicircles meet here; they are important as well, but that's beyond my point right now), then the sequence of derivatives need not be convergent.
A: $\lim \limits_{n\to\infty}{2^n*Pi/2^n}$
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).
A: The arc length should also be $\pi$. The curve "converges" to a straight line, but is actually an infinite union of semicircles of 0 length. You cannot approximate the arc length like that, it is similar to the fallacious proofs that $\pi=4$. In fact, there are $2^n$ semicircles each with diamater $\frac{1}{2^n}$, then the arc length is $2^n\pi\frac{1}{2^n}=\pi$.
