Consider the following sequence of piece-wise smooth functions $f_n:[0,2]\rightarrow \mathbb{R}$ (below drawn first two functions by their graphs, and is contunued then in natural way.)
The function $f:[0,2]\rightarrow \mathbb{R}$, $f(x)=0$ for all $x$ is the limit of this sequence.
Q. 1 Note that the lengths of graph of each $f_n$ is $2\sqrt{2}$, but the length of limiting function is not $2\sqrt{2}$; why this happens? (Such problem appears as a puzzle shown below picture in some book, but I was looking it through a sequence of functions, it convergence etc.)
Here sequence $f_n$ converges to $f$ uniformly; but still the length of their graphs do not converges to length of graph of limit of $f_n$. This raises following question:
Q If $f_n:[a,b]\rightarrow \mathbb{R}$ is a sequence of piecewise smmoth functions, and converging uniformly to $f:[a,b]\rightarrow \mathbb{R}$, then under what more conditions on $f_n$, we can guarantee the convergence of lengths of $f_n$ to length of $f$?
Let $L(\,f)$ denote the length of the graph \begin{align} \Gamma_f = \{t+if(t) \in \mathbb{C} : t \in [0, 2]\} \,. \end{align}
Let $\{\,f_n\}_{n=1}^\infty$ be a sequence of real-valued continuous functions that converge uniformly on $[0, 2]$ to the function $f:=\lim\limits_{n \to \infty}\,f_n$. We have that $\,L(\,f) = \lim\limits_{n \to \infty} L(\,f_n)$ holds if and only if the sequence of functions $\{\,f_n\}_{n=1}^\infty$ forms on $[0,2]$ an absolutely equicontinuous family such that $f^{'}_n$ converges in measure to $f'$.
I can't give a formal proof of the preceding statement. Though I took the liberty of coding up a worse counterexample sequence $\{g_n\}_{n=1}^\infty$ that I found more insightful \begin{align}f_n(x) := \begin{cases} x-\frac{2k-2}{n} & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right], \\ \frac{2k}{n}-x& \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] \end{cases} && (k=1,\ldots,n) \end{align}
\begin{equation} \end{equation}
\begin{align}g_n(x) := \begin{cases} n^{n-1}\left(x-\frac{2k-2}{n}\right)^n & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right], \\ n^{n-1}\left(\frac{2k}{n}-x\right)^n & \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] \end{cases} && (k=1,\ldots,n) \end{align}
\begin{equation} \begin{aligned} & L(g_j) =2j \int_0^{1/j} (1+j^{2j}t^{2j-2})^{1/2} dt \\& L(g_j) \leq L(g_{j+1})\end{aligned} \: \;\: \;(\,j=1,2,\ldots) \end{equation}
Note: $g_1(x)=f_1(x)$ for all $x \in [0,2].$
I should mention that there is a sequence of functions similar to $\{\,f_n\}_{n=1}^\infty$ $(*)$ described in a comment by ThePortakal, here: Sequence of functions that converges uniformly in $R$, derivatives does not converge punctually at any point of $R$.
$(*)$ The family of functions described by ThePortakal is absolutely equicontinuous on $[0,2]$, and the graphs corresponding to the functions in this family will have constant length $\pi$ on $[0,2]$.
I have decided to add a picture of the work that went into my original answer (excuse the notation abuse with $\,f_n$ on the right-hand side)
Along with Beginner's interesting question here and ThePortakal's comment in another thread, this Dover book was helpful in motivating the sequence $\{g_n\}_{n=1}^\infty$.