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At a major axis ratio of 1:15, you're already well in the territory where a double-elliptic transfer is more efficient. The reason is actually exactly what was described in this video.
I imagine if you tried to overshoot by an infinite amount, you'd reach a point where you are so far away that the gravity of the object you are orbiting can't pull you back faster than the universe is expanding. So there must be some finite limit imposed by the fact that the universe is expanding.
The other problem with bi-elliptical transfers (and going to infinity) is that eventually you're going to hit the gravity well of another body and that'll screw up all your planning.
There's also the fact that if you swing your orbit out far enough, you'll eventually find a second gravitation well you can swing off of for even more fuel savings.
The really fun thing is that since everything is symmetric, it's also more efficient to use a bi-elliptic transfer to *lower* your orbit. The first step is shooting yourself into an even higher elliptical orbit. Going up in order to go down.
Two approximations in this video:
First, the graphs use Δv as the cost, which is definitely not the same as fuel. In fact, a small saving in Δv can result in a huge saving in fuel.
Second, an infinite bi-elliptic transfer is, technically, a bi-parabolic transfer.
2:52 I have actually used this in KSP before lmao. Just barely didn't have enough fuel to return to kerbin, so I ejected myself slightly further out and came back with 12m/s to spare.
6:30 - There are also other mass out there, so if you go too far out, your elliptical orbits might have its trajectory altered by the gravity well of some other astronomical object and cause you to never reach the theoretical recircularization point. Even if you're overshooting by a large-but-finite distance.
theres another problem - when you go so far out you must be extremely precise on how much of a burn you do, so at these insane distances of overshooting thousands of times farther you must be very precise, or you will end up needing to make a course-correcting burn, and at the millions of times overshooting its simply impossible to be so precise so you will be needing to make a course-correcting burn
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First, the graphs use Δv as the cost, which is definitely not the same as fuel. In fact, a small saving in Δv can result in a huge saving in fuel.
Second, an infinite bi-elliptic transfer is, technically, a bi-parabolic transfer.