# Untitled Page 16

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## 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*=37*d*, then *d*/*e*=1/37 is finite; but if
*e*=*d*^{2}, then *d*/*e* is infinite; and if *d*=*e*^{2}, 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,

### Example 14

◊ 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.