How does approximating the circumference of a circle using regular n-gons work, if each regular n-gon isn't continuously differentiable? For a function $f$  to have an arc length in the interval [a,b], $f'$ must exist in the interval [a,b], so $f \in C^1[a,b]$. If we want to approximate the arc length of $f$ with ${f_n}$ we need that $f'_n$ converges uniformly to $f'$ and $f_n$ converges to $f$ for some $x_o \in [a,b]$
If $f_n$ is the top half of a n-gon (and each n-gon is a combination of line segments) and $f$ is the semicircle then how does $f'_n$ converge uniformly to $f'$ if $f_n \notin C^1[a,b]$ because of the cusps in each n-gon
Does the arc length still converge because even though $f'_n$ doesnt exist in finitely many values of $x \in [a,b]$, $\int\sqrt{1+f_n^{'2}}$ $dx$ still exists?
 A: The definition of arc length of a curve $\mathcal{C}$ given by $$x = f(t), y = g(t)\tag{1}$$ where $f, g$ are continuous on some closed interval $[a, b]$ is given in a manner similar to that of Riemann integration. We take a partition $$P = \{t_{0} = a, t_{1}, t_{2}, \ldots, t_{n} = b\}$$ of the interval $[a, b]$ and consider the points $P_{k} = (f(t_{k}), g(t_{k}))$ on the curve $\mathcal {C}$. The length of the polygonal arc $P_{0}P_{1}\cdots P_{n}$ is given by the sum $$L(P) = \sum_{k = 1}^{n}P_{k - 1}P_{k} = \sum_{k = 1}^{n}\sqrt{\{f(t_{k}) - f(t_{k - 1})\}^{2} + \{g(t_{k}) - g(t_{k - 1})\}^{2}}\tag{2}$$ and it is clear that $L(P)$ is non-decreasing as we add points in the partition $P$.

If the set of all such sums like $L(P)$ for all partitions $P$ of $[a, b]$ is bounded then we say that the curve in $(1)$ has an arc-length $L$ given by $$L = \sup \,\{L(P)\mid P\text{ is a partition of }[a, b]\}\tag{3}$$ Thus the definition of arc-length itself is based on the usage of polygonal lines connecting points on the curve and then taking the limit of length of this polygonal arc. The same applies to the circle also and taking the polygons to be regular (or irregular) does not create any issue.

It is not difficult to prove that the curve given by $(1)$ possesses an arc-length (i.e. the supremum $L$ in $(3)$ exists) if and only if both the functions $f, g$ are of bounded variation in $[a, b]$. Further, a function is of bounded variation if and only if it can be expressed as the difference of two increasing functions.

Sometimes the curve is given by means of simpler equation like $x = t, y = f(t)$ so that the curve is graph of $y = f(x)$ (this is the case you are interested in) and then it possesses an arc-length if and only if $f$ is of bounded variation on $[a, b]$. Note that if $f$ possesses a bounded derivative $f'$ then it implies that $f$ is of bounded variation, but it is not necessary that $f'$ must be bounded for $f$ to be of bounded variation.
Note further that if $f$ is of bounded variation then it is possible that $f'$ may not exist at some points in $[a, b]$. Thus from the view-point of existence of arc-length the right criterion is not related to derivative. However the calculation of arc-length is convenient if $f'$ is integrable and we have the formula $$L = \int_{a}^{b}\sqrt{1 + f'^{2}(x)}\,dx$$

The concept of functions of bounded variation and their link with arc length of a curve is well explained in Tom Apostol's Mathematical Analysis and I have extracted some of this material in "series of blog posts on monotone functions and functions of bounded variation" starting with this post. In case you are just interested in the details of arc length of a curve you can straightaway jump to this blog post in the above series.
