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