Do all the curves become straight line in limit of their length to zero.
I know $$ \triangle L=\sqrt{(\triangle x)^2 + (\triangle y)^2}$$
And when $ \triangle L ,\triangle x , \triangle y$ $\longrightarrow 0$
$$ d L=\sqrt{(d x)^2 + (d y)^2}=dx \sqrt{....}$$
which $\sqrt{\cdot}$ is defined according to our curve equation.
So, does the curve becomes straight line when $\triangle L \longrightarrow 0$ ?
All curves which are everywhere differentiable become straight lines (in the sense you are trying to describe, which can in fact be made into a rigorous definition) as the diameter of the region examined goes to zero.
And the proof of that does go along the lines of your reasoning.
However, curves which are not everywhere differentiable do not have that property: Consider $y=|x|$ near $x=0$.
You might think that you can have a fractal curve which is everywhere differentiable, but which does not become straight lines in sufficiently small regions. However, fractal curves are not everywhere differentiable.
Maybe it can be formulated this way.
Take a function $f(x)$. Also, consider points $x_0$ and $x_0+dx$. The line that goes through the points is
$y=\alpha x-\alpha x_0 +f(x_0)$
where $\alpha=\frac{f(x_0+dx)-f(x_0)}{dx}$
Then, we can define when a curve between the points tends to a line, as below
$\forall\epsilon>0 \exists dx>0|\forall x\in (x_0,x_0+dx) \rightarrow |f(x)-\alpha x+\alpha x_0 -f(x_0)|<\epsilon$
Using triangle inequality, we can have
$|f(x)-\alpha x+\alpha x_0 -f(x_0)|\leq |f(x)-f(x_0)|+|\alpha||x-x_0|$
Assuming $|\alpha|<\infty$ , we need to make sure that
$lim_{x\rightarrow x_0}|f(x)-f(x_0)|=0$
So, I think, for a function $f(x)$, the requirement is to have
$lim_{x\rightarrow x_0}f(x)=f(x_0)$