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).

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 $2^n\pi\frac{1}{2^n}=\pi$.
