Show that the arc length is a limit of lengths of inscribed polygons. In Do Carmo's book (page 11), asked to find given $\epsilon>0$, there exists  $\delta>0$ such that if $|P|<\delta$ then $\left|\int_a^b |\alpha'(t)|~dt-l(\alpha,P)\right|<\epsilon$.
Here $P$ is a partition with $|P|=\max(t_i,t_{i-1}),i=1, \dots, n$, $\alpha:I \to \mathbb{R}^3$ for an interval $I$,$l(\alpha, P)= \sum_{i=1}^n|\alpha(t_i)-\alpha(t_{i-1})|$.
Any ideas to prove this?
 A: Fix $\varepsilon=2\epsilon/(b-a)$.
In each interval $[t_i,t_{i-1}]$ we have $$\left|\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)\right|=\int_{t_{i-1}}^{t_{i}}\left|\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|dt$$
so that
$$
\begin{align}
\left|\int_{t_{i-1}}^{t_{i}}\left|\alpha'\left(t\right)\right|dt-\left|\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)\right|\right|
& =
\left|\int_{t_{i-1}}^{t_{i}}\left[\left|\alpha'\left(t\right)\right|-\left|\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|\right]dt\right|
\\ & \leq 
\int_{t_{i-1}}^{t_{i}}\left|\left|\alpha'\left(t\right)\right|-\left|\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|\right|dt
\\ & \leq 
\int_{t_{i-1}}^{t_{i}}\left|\alpha'\left(t\right)-\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|dt
\end{align}
$$
by the triangle inequality.
We can write $$\left|\alpha'\left(t\right)-\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|\leq\left|\alpha'\left(t\right)-\alpha'\left(t_{i}\right)\right|+\left|\alpha'\left(t_{i}\right)-\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|$$ so that by continuity of $\alpha'$ there is a $\delta_1>0$ such that $\left|\alpha'\left(t\right)-\alpha'\left(t_{i}\right)\right|<\varepsilon$ when $\left|t-t_i\right|<\delta_1$ and, since $\alpha'$ exist, by definition there is a $\delta_2>0$ such that $\left|\alpha'\left(t_{i}\right)-\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|<\varepsilon$ when $\left|t_{i}-t_{i-1}\right|<\delta_2$.
Put $\delta = \min(\delta_1,\delta_2)$ and $|P|<\delta$. Then since $\left|t-t_i\right|\leq \left|t_{i}-t_{i-1}\right|<\delta$ when $t \in [t_{i-1},t_i]$ for this $\delta$ both inequalities holds so that $$\left|\alpha'\left(t\right)-\alpha'\left(t_{i}\right)\right|+\left|\alpha'\left(t_{i}\right)-\frac{\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)}{t_{i}-t_{i-1}}\right|<2\varepsilon$$ and $\left|\int_{t_{i-1}}^{t_{i}}\left|\alpha'\left(t\right)\right|dt-\left|\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)\right|\right|<2\varepsilon(t_i-t_{i-1})$.
Therefore $$
\begin{align}
\left|\int_{a}^{b}\left|\alpha'\left(t\right)\right|dt-\sum_{i=1}^{n}\left|\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)\right|\right|
& = 
\left|\sum_{i=1}^{n}\left[\int_{t_{i-1}}^{t_{i}}\left|\alpha'\left(t\right)\right|dt-\left|\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)\right|\right]\right|
\\ & \leq
\sum_{i=1}^{n}\left|\int_{t_{i-1}}^{t_{i}}\left|\alpha'\left(t\right)\right|dt-\left|\alpha\left(t_{i}\right)-\alpha\left(t_{i-1}\right)\right|\right|
\\ & \leq
\sum_{i=1}^{n}2\varepsilon\left(t_{i}-t_{i-1}\right)
\\ & = 
2\varepsilon\left(b-a\right)
\\ & =
\epsilon
\end{align}
$$
as claimed.
