In computer science, there's a semi-joking saying that $\log n < 50$ for all $n$. Of course this isn't true – as you say the logarithm is unbounded. But what is true is that it grows so slowly that, if you can only put in quantities like memory size, mass of some material, time etc., then the logarithm is “effectively bounded”, because it is (literally) exponentially expensive to grow it any further.
At first glance, this doesn't seem to apply to your question, because you're only shrinking something to zero. In this case, the cost is not of supplying an anfeasible amount of something, but of making the circuitry behave so nearly ideal that the tiny residual voltage can still be measured. In particular, you need to reduce the noise; if there's more noise than signal voltage you can't measure the latter. Experimentally, this pretty much comes down to cooling down the measurement amplifiers (in nuclear physics, often detectors need liquid nitrogen cooling simply for the preamplifiers to work more accurately!), and reducing interference by electromagnetic shielding.
Either that or, relevant here: you can also reduce noise by introducing large capacitors into the circuit (in this case, there is still noise, but it's “filtered out” again). That's often avoided because, obviously, it makes the circuit's response slow; in this particular experiment it would mean that we continue to measure a voltage difference, but we can't really tell whether it is from the capacitor we're actually interested in, or from the one we just used in order to make the measurement more “accurate”.