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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=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,

\begin{align} \frac{.001}{.037} &= \frac{1}{37} \frac{.001}{.000001} \\ &= 1000 \frac{.000001}{.001} \\ &= .001 . \end{align}

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,

\begin{align} 1000-500 &= 500 \\ 1001-1000 &= 1 . \end{align}

Example 45

◊ If H is a positive infinite number, is \(\sqrt{H+1}-\sqrt{H-1}\) finite, infinite, infinitesimal, or indeterminate?

◊ Trying it with a finite, big number, we have

\begin{align} & \sqrt{1000001}-\sqrt{999999} \\ &= 1.00000000020373\times 10^{-3} , \end{align}

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 \(\sqrt{H}(\sqrt{1+1/H}-\sqrt{1-1/H})\). Since the derivative of the square root function \(\sqrt{x}\) evaluated at x=1 is 1/2, we can approximate this as

\begin{align} \sqrt{H} & \left[1+\frac{1}{2H}+...-\left(1-\frac{1}{2H}+...\right)\right] \\ &= \sqrt{H}\left[\frac{1}{H}+...\right] \\ &= \frac{1}{\sqrt{H}} , \end{align}

which is infinitesimal.