Understanding limits of length Also, I want to talk about what is technically a joke:

This is a joke, and I know why it is wrong. There's another thread about it. 
However, looking through a book on multivariable calculus, it also exemplifies what sometimes confuses me about limits. How do I, for instance, know that the formula for the arc length of a parametric curve gives us the length of the parametric curve? It's proved in terms of limits. I think I know why the joke in the picture above is wrong, but I do not know why the formula for the arc length is correct
The answers for why the picture above is incorrect say a lot about the Euclidean metric versus the Manhattan metric. That's okay. I understand why the Euclidean metric gives us a more correct measure of length when we are talking about straight lines. However, once limits get involved, I cannot rigorously say why the proof here is more accurate than an approach more like the one in the picture above.
I have had real and complex analysis. I understand Lebesgue integration, and I think I understand Sylow's theorems. However, I still have a problem "connecting" calculus with basic concepts from geometry.
 A: This is an interesting question. I think the key to it is to delve into the proof that two expressions for the length of a smoothly parameterized curve $\gamma : [a,b] \to \mathbb{R}^n$ give the same answer. The key thing to exploit is that the vector valued derivative $\gamma'(t)$ is uniformly continuous, which obviously fails for the Manhattan metric approximating curve.
The first expression for length is the one from calculus:


*

*$\text{Length}(\gamma) = \int_a^b \bigl| \gamma'(t) \bigr| \, dt$


and the other from metric space theory: 


*$\text{Length}(\gamma)$ equals the supremum, over all partitions $a=t_0 < t_1 < ... < t_N = b$, of the length of the secant segment approximation $\sum_{n=1}^N \bigl|\gamma(t_n)-\gamma(t_{n-1})\bigr|$


Let me try to isolate the key issue of the proof of equality of 1 and 2 with a simple problem. 
Suppose that I have a continuously differentiable function $f : [a,b] \to \mathbb{R}$ such that $f(a)=f(b)$, and hence the slope of the secant line between $(a,f(a))$ and $(b,f(b))$ equals zero. Suppose also that we have an upper bound $|f'(x)| \le B$ for all $x \in [a,b]$, and we imagine that $B$ is very close to zero. Consider the calculus expression the length of the graph $\Gamma = \{(x,f(x) \,|\, x \in [a,b]\}$
$$\text{Length}(\Gamma) = \int_a^b \sqrt{1 + f'(x)^2} \, dx \, 
$$
What can be said about the value of this expression?
Well, we have an estimate for the integrand of the form
$$1 \le \sqrt{1 + f'(x)^2} \le \sqrt{1+B^2}
$$
which leads to a length estimate
$$\text{Length}[a,b] \le \text{Length}(\Gamma) \le \sqrt{1+B^2} \, \text{Length}[a,b]
$$
which leads to a length ratio estimate
$$1 \le \frac{\text{Length}(\Gamma)}{\text{Length}[a,b]} \le \sqrt{1+B^2}
$$
So as the bound $B$ gets closer and closer to zero, the length ratio gets closer and closer to $1$.
One can generalize this to the case that $f(a) \ne f(b)$, where the nature of the slope bound should be expressed in terms of the angle of the tangent line at each point compared to the angle of the secant line between $(a,f(a))$ and $(b,f(b)$, rather than in terms of slopes. One again obtains a uniformly bounded ratio of lengths, expressed in terms of a bound in the differences between angles.
Getting back to why 1 and 2 give the same result, the point is that if one chooses the partition $a=t_0<t_1<...<t_N=b$ to be very fine, uniform continuity of $\gamma'(t)$ guarantees that on the very short subinterval $[t_{n-1},t_n]$ the angle of $\gamma'(t)$ is very close to the angle of the secant segment approximation that connects the two points $\gamma(t_{n-1})$ and $\gamma(t_n)$ (where "very close" means that the bound is uniform over the segments). Therefore the integral expression for $n^{\text{th}}$ arc of $\gamma$, namely $\text{Length}(\gamma[t_{n-1},t_n])$, and the $n^{\text{th}}$ secant segment length, namely $\bigl| \gamma(t_n)-\gamma(t_{n-1})\bigr|$, have a ratio that is very close to $1$. Taking the sum from $n=1$ to $N$ we see that $1$ and $2$ have ratio that is very close to $1$. The ratio gets close and closer as the partition gets finer and finer, and hence 1 and 2 give the same answer. 
A: Assume that $Y_n$ is a sequence of broken curves in OP. Consider induced length metric $d_n,\ d$ on $Y_n,\ X=S^1$. Note that embeddings $Y_n,\ X\rightarrow {\bf R}^2$ are not isometric. So we can not conclude that $(Y_1,d_1)=(Y_n,d_n)$ is isometric to $ (X,d)$
A: The sequence of approximating curves converge to the circle in many ways (geometrically, visually, max(distance),...). That is however not a guarantee that a function(al) on curves (like length for example) will be suitably continuous. You need continuity of the function(al) to infer that the sequence of lengths converge to the length of the circle.
It is the other direction that is most often relevant in practice, i.e. for approximation purposes. Given a function L on curves, work with convergence notions that make L continuous.
