The Wikipedia article on arc length of a smooth curve says the following of a continuously differentiable function $f:[a\,..b]\rightarrow \mathbb R^n$ ($[a\,..b]$ having been subdivided into a sequence of $t_i$ where $t_i=a+i\dfrac{(b-a)}N$ for $i=0,1,...,N$ and $\Delta t=|t_i-t_{i-1}|$):
[...] definition of the derivative as a limit implies that there is a positive real function ${\displaystyle \delta (\epsilon )}$ of positive real ${\displaystyle \epsilon }$ such that ${\displaystyle \Delta t<\delta (\epsilon )}$ implies ${\displaystyle \left|{\bigg |}{\frac {f(t_{i})-f(t_{i-1})}{\Delta t}}{\bigg |}-{\Big |}f'(t_{i}){\Big |}\right|<\epsilon .}$ This means
${\displaystyle \sum _{i=1}^{N}\left|{\frac {f(t_{i})-f(t_{i-1})}{\Delta t}}\right|\Delta t-\sum _{i=1}^{N}{\Big |}f'(t_{i}){\Big |}\Delta t}$ has absolute value less than ${\displaystyle \epsilon (b-a)}$ for ${\displaystyle N>(b-a)/\delta (\epsilon ).}$
So far I understand why differentiability of $f$ implies the statement in the first part of the quote (so everything before "This means"). I don't understand how this statement implies the latter part. How, from the former can we deduce the latter?
Using the triangle inequality,
$$\left|\sum_{i=1}^N \left(\left|\frac{f(t_i)-f(t_{i-1})}{\Delta t}\right|\Delta t-|f'(t_i)|\Delta t\right)\right|=\left|\sum_{i=1}^N \left(\left|\frac{f(t_i)-f(t_{i-1})}{\Delta t}\right|-|f'(t_i)|\right)\Delta t\right|\leq \\ \leq \sum_{i=1}^N\left|\left|\frac{f(t_i)-f(t_{i-1})}{\Delta t}\right|-|f'(t_i)|\right|\Delta t \leq \sum_{i=1}^N \varepsilon \cdot \Delta t \leq N\varepsilon \sum_{i=1}^N \Delta t = \varepsilon (b-a)$$
the last equality being true because $$\sum_{i=1}^N \Delta t= (t_1-t_0)+(t_2-t_1)+\dots+(t_N-t_{N-1})=t_N-t_0=b-a$$