Here is a mathematical game:
Let $x > 0$ be a real number. We know nothing about $x$, but we are given a black box that inputs any real number $y$ and tells us if $y$ is greater, less than or equal to $x$.
We are tasked to find the best possible approximation for $x$, but we are allowed to call the black box no more than $n$ times.
To illustrate, here’s one possible (rather naive) solution:
We define a sequence of rational numbers $(p_i/q_i)_{1\leq i \leq n}$ as follows. First set $(p_1,q_1)=(1,1)$. For $i=1,2,\ldots,n$, we input $p_i/q_i$ into the black box, and subsequently define:
$$(p_{i+1},q_{i+1}) = \begin{cases} (p_i+1,q_i) &\mbox{if } p_n/q_n <x \\ (p_i,q_i+1) & \mbox{if } p_n/q_n > x \\ (p_i,q_i) & \mbox{if } p_n/q_n =x \end{cases}$$
The above sequence of fractions surely converges to $x$ as $n\to \infty$, and what's more the error term $\varepsilon_n :=|x - p_n/q_n|$ decays in $O(1/n).$
Is there an algorithm such that the error term decays quadratically? Cubically? Exponentially?
Suppose Alice has an algorithm that could address the seemingly easier problem of just managing this for $x\in(0,\,1)$. Bob's favourite strictly increasing function $f$ from $(0,\,1)$ to $\Bbb R^+$ (it might for example be $t\mapsto t/(1-t)$) allows him to estimate $x$ by using Alice's technique to estimate $f^{-1}(x)$. If Alice's method would try $y$ as such an estimate, Bob just uses $f(y)$ as an estimate of $x$ instead. Thus we can focus instead on how Alice solves her bounded problem.
Here's something Alice could try: start with $0.5$ so we halve the range of values of $x$, go to the middle of that and so on. For example, if $x=0.83624$ the estimates are $0.5,\,0.75,\,0.875,\,0.8125,\,0.94375,\,\cdots$. This is called a binary search. Our error in $x$ is $O(2^{-n})$. Going back to Bob's harder problem (i.e. the one you asked about), the error in estimating $x\in\Bbb R^+$ is by the chain rule approximately $f'f^{-1}(x)$ times the error in estimating $f^{-1}(x)$. Again, the error is $O(2^{-n})$. If you pick $f$ not to flatten too fast, the $f'f^{-1}(x)$ factor won't be too bad. Fo example the rational function I suggested above would be a much better choice of $f$ than $t\mapsto \operatorname{artanh} t$.
Can we do better than a binary search? Not really. Each yes/no question of the black box gives us $1$ bit of information, so can only halve the measure of values of $x$ consistent with our information. The advantage of Alice's bounded problem is it lets us work with finite-measure sets.