In practice you might also want to use a O(n^2) algorithm like insertion sort under some threshold.
Sure, but the author didn't argue that the simpler algorithm would be faster for 5 items, which would indeed make sense.
Instead, the author argued that it's OK to use the simpler algorithm for less than 5 items because 5 is a constant and therefore the simpler algorithm runs in constant time, hence my point that you could use the same argument to say that 2^140 (or 2^256) could just as well be used as the cut-off point and similarly argue that the simpler algorithm runs in constant time for all arrays than can be represented on a real-world computer, therefore obviating the need for the more complex algorithm (which obviously makes no sense).
In the article n was set to 5. All of those arrays (except maybe 1) have exactly 5 elements. There is no variance (and even if there was, it would be tiny, there is no point in talking about limits of 5-element sequences).
No, the code was:
# If there are < 5 items, just return the median
if len(l) < 5:
return nlogn_median(l)
So your point is: not all constants are created equal. Which circles all the way back to my original point that this argument is pretty funny :)