3.4 Another perspective on indeterminate forms
An expression like 0/0, called an indeterminate form, can be thought of in a different way in terms of infinitesimals. Suppose I tell you I have two infinitesimal numbers d and e in my pocket, and I ask you whether d/e is finite, infinite, or infinitesimal. You can't tell, because d and e might not be infinitesimals of the same order of magnitude. For instance, if e=37d, then d/e=1/37 is finite; but if e=d2, then d/e is infinite; and if d=e2, then d/e is infinitesimal. Acting this out with numbers that are small but not infinitesimal,
On the other hand, suppose I tell you I have an infinitesimal number d and a finite number x, and I ask you to speculate about d/x. You know for sure that it's going to be infinitesimal. Likewise, you can be sure that x/d is infinite. These aren't indeterminate forms.
We can do something similar with infinite numbers. If H and K are both infinite, then H-K is indeterminate. It could be infinite, for example, if H was positive infinite and K=H/2. On the other hand, it could be finite if H=K+1. Acting this out with big but finite numbers,
◊ If H is a positive infinite number, is √H+1-√H-1 finite, infinite, infinitesimal, or indeterminate?
◊ Trying it with a finite, big number, we have
which is clearly a wannabe infinitesimal. We can verify the result using Inf:
: H=1/d d^-1 : sqrt(H+1)-sqrt(H-1) d^1/2+0.125d^5/2+...
For convenience, the first line of input defines an infinite number H in terms of the calculator's built-in infinitesimal d. The result has only positive powers of d, so it's clearly infinitesimal.
More rigorously, we can rewrite the expression as √H(√1+1/H-√1-1/H). Since the derivative of the square root function √x evaluated at x=1 is 1/2, we can approximate this as
which is infinitesimal.