Convergence of the polygonal approximation of a differentiable 3-space curve The following is a question taken from do Carmo's Differential Geometry of Curves and Surfaces. I believe I have an alternative method of proof to the one he suggests in the hints, but I would just like to confirm its validity, and reach some clarity on where it might fail. I'm actually fairly new to $\epsilon$-$\delta$ proofs, so please excuse any obvious blunders:
Q: Let $ \alpha \colon I \to \mathbb{R}^3 $ be a differentiable curve and let $[a,b] \subset I$ be a closed interval. For every partition $$ a=t_0 < t_1 < \ldots <t_n = b $$ of $[a,b]$, consider the sum $ \sum_{t=1}^{n} \lvert \alpha(t_i) - \alpha(t_{i-1}) \rvert = l(\alpha,P)$, where $P$ stands for the given partition. The norm $\lvert P \rvert$ of a partition $P$ is defined as $$ \lvert P \rvert = \max(t_i - t_{i-1}), \ i=1,\ldots,n. $$ Prove that given $\epsilon > 0$ there exists $\delta > 0$ such that if $ \lvert P \rvert < \delta $ then $$  \left\lvert\int_{a}^{b} \lvert \alpha'(t) \rvert dt - l(\alpha,P) \right\rvert < \epsilon $$ A: This is equivalent to showing that given any $\epsilon>0$, there exists a $\delta>0$ such that $$ \max(t_i - t_{i-1}) < \delta  \implies \left\lvert\int_{a}^{b} \lvert \alpha'(t) \rvert dt - \sum_{i=1}^{n} \lvert \alpha(t_i) - \alpha(t_{i-1}) \rvert \right\rvert < \epsilon. $$
Since $\alpha$ is differentiable, this implies the component functions $ x(t), y(t), z(t) $ are differentiable. Applying the mean value theorem to the first component (the other components are treated similarly) we get $$ x'(c_1) = \frac{x(t_i) - x(t_{i-1})}{t_i - t_{i-1}}, \ \text{for some} \ c_1 \in [t_{i-1},t_{i}] $$ and since $ t_i - t_{i-1} < \delta $ we have $$ x(t_i) - x(t_{i-1}) < \delta x'(c_1), $$ $$ y(t_i) - y(t_{i-1}) < \delta y'(c_2), $$ $$ z(t_i) - z(t_{i-1}) < \delta z'(c_3). $$ So we have $$ \lvert \alpha(t_i) - \alpha(t_{i-1}) \rvert = \sqrt{(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1}))^2} $$ $$ < \delta \sqrt{(x'(c_1))^2 + (y'(c_2))^2 + (z'(c_3))^2} = \delta A_{i-1}. $$ Setting $A=\max{A_{i-1}}$, we see that $$ \sum_{i=1}^{n} \lvert \alpha(t_i) - \alpha(t_{i-1}) \rvert < n \delta A. $$ Now $$\left\lvert \int_{a}^{b} \lvert \alpha'(t) \rvert dt  \right\rvert \le \int_{a}^{b} \lvert \alpha'(t) \rvert dt = \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} \lvert \alpha'(t) \rvert dt = \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt $$ and by the definite integral mean value theorem there exists a $d_{i-1} \in [t_{i-1},t_i]$ such that $$ \int_{t_{i-1}}^{t_i} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt = (t_i - t_{i-1}) \sqrt{(x'(d_{i-1}))^2 + (y'(d_{i-1}))^2 + (z'(d_{i-1}))^2} $$ $$ = (t_i - t_{i-1}) \lvert \alpha'(d_{i-1}) \rvert < \delta \alpha'(d_{i-1}). $$ Setting $B = \max{\lvert \alpha'(d_{i-1}) \rvert}$, we see that $$ \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt < n\delta B. $$ Putting this all together and applying a triangle inequality we get $$ \left\lvert\int_{a}^{b} \lvert \alpha'(t) \rvert dt - \sum_{i=1}^{n} \lvert \alpha(t_i) - \alpha(t_{i-1}) \rvert \right\rvert \le \left\lvert\int_{a}^{b} \lvert \alpha'(t) \rvert dt \right\rvert + \left\rvert \sum_{i=1}^{n} \lvert \alpha(t_i) - \alpha(t_{i-1}) \rvert \right\rvert < \delta (nA + nB).$$ Set $ \delta = \frac{\epsilon}{n(A+B)} $ to get the result. $ \ \blacksquare $
