I was reading about the Cantor function (The Devil's Staircase) here and based on the picture and the description of the function, it seems to me that when $x \rightarrow 1$, $$1-c(x)=o(1-x).$$
Is it true? Is in this case the rate of $1-c(x)$ going to $0$ as $x \rightarrow 1$ well defined?
I am a bit confused by your o-notation. But I guess that you are interested in the left-sided derivative at the point 1 -- it is infinite. Described by limits: $$\lim_{x\to 1} \frac{1-c(x)}{1-x} = \infty.$$
From now on, I will focus on the behavior at the point 0 -- it is more handy and the Cantor function is symmetric by point [0.5, 0.5].
Take the two sequences $a_n = 3^{-n}$, $b_n = 2\cdot3^{-n}$ approaching to zero. They converges to zero and they have the same values in the Cantor function: $c(a_n) = c(b_n) = 2^{-n}$. We can describe the behavior of $c$ on the sequence $a_n$ by function $f(x)=x^{\log_3(2)}$ and on the sequence $b_n$ by function $g(x) = (\frac x2)^{\log_3(2)} = 2^{-\log_3(2)}\cdot f(x)$.
These functions are actually bounds for the Cantor function, given that we can say that $c\asymp x^{\log_3(2)}$, using the asymptotic notation 6..
I could prove the fact that $f,g$ are the bounds, but it is rather technical, so I hope it won't be necessary :-)