I'm trying to calculate the order of convergence for the bisection method. Given a continuous function $f\in C[a, b]$ such that $f(a)f(b) < 0$, I want to show that: $$ \lim_{{n \to \infty}} \frac{|p_{n+1} - p|}{|p_n - p|} = \frac{1}{2} $$
What I tried to do
I know that $|p_n - p| \leq \frac{b-a}{2^n}$, and so we also have: $|p_{n+1} - p| \leq \frac{b-a}{2^{n+1}}$
Thus, the numerator $|p_{n+1} - p|$ can be bounded by $\frac{b-a}{2^{n+1}}$, but to find an upper bound for the limit expression, I need to enlarge the limit expression, which is a quotient. After enlarging the numerator, I am left with trying to find a lower bound for the denominator. But I don't see any lower limit for this expression and we can possibly intercept the root with one of our iteration points.
The difficulty
I think I am missing something basic to Numerical Analysis error analysis or order of convergence, because even if we assume that root does not intercept the target function root, how can we find a lower bound for the denominator? I don't think this quotient limit even exist. And even if we conceptualize this limit as a series of lengths, this limit is not a limit in the ordinary sense, because we are enlarging both the numerator and denomintor. So what is this limit, and how does this relate to order of convergence?
Other approaches
One approach I considered, which is not rigorous and definitly "smells", is using the concept of order of convergence, while fixing the relative series and constant used:
Since $|p_n - p| \leq \frac{b-a}{2^n}$, we have $p_n \leq p + K \left(\frac{1}{2^n}\right)$ for $K=b-a$. Thus, we have $p_n = p+O\left(\frac{1}{2^n}\right)$. Similarly, since $|p_{n+1} - p| \leq \frac{b-a}{2^{n+1}}$, we have $p_n \leq p + \frac{1}{2}K \left(\frac{1}{2^n}\right)$, also for $K=b-a$. Thus, we have: $p_{n+1} = p+\frac{1}{2}O\left(\frac{1}{2^n}\right)$. After expressing both $p_n, p_{n+1}$ in terms of rate of convergence with respect to the same series $\frac{1}{2^n}$, and considering we have the same constant $K$, for both expressions, maybe we do the following:
$$ \frac{|p_{n+1} - p|}{|p_n - p|} = \frac{1}{2}\left(\frac{O\left(\frac{1}{2^n}\right)}{O\left(\frac{1}{2^n}\right)}\right) = \frac{1}{2} $$
But this "smells", and I am only hiding the difficulty, which is the limit calculation for the order of convergence.
Related questions
1: Convergence of Bisection method
The accepted answer there states that $\frac{\epsilon_{n+1}}{\epsilon_n}=\frac{1}{2}$, but I think that $\epsilon_n=|p_n-p|$, and so, again, it seems that absolute error upper bounds were used instead of the error bounds for the limit.
2: Rigrous proof of the rate and order of convergence of bisection method
This seems close to what I am asking, but I could not understand the answer (and really most of the question), and the meaning (and difficulty) of the original limit calculation were not discussed.
