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08 August 2003

Extreme Limits

by George E. Hrabovsky, President, MAST

A Continuation

Last week we explored the ideas of arithmetic of limits.  Today we will examine the limits involving infinity and zero.

Infinite Limits

So far we understand that in the case of a limit as x  x_0 the difference | f(x) - L | becomes arbitrarily small.  So, it is reasonable to ask the question, "What happens when the difference becomes arbitrarily large?"

If we use the traditional symbol M to represent an arbitrarily large number, and we have some function f(x), we can define this function to be on a deleted neighborhood of some point x_0.  If, as x  x_0 we have any number M>0 no matter how large such that there exists some sufficiently small number δ = δ(M) >0 such that for every value of x we know that | f(x) | > M such that 0 < | x - x_0 | < δ, then we say that f(x) approaches infinity.  We can write this f(x)  ∞ as x  x_0.  Another way of writing this is,

Underscript[lim, xx_0] f(x) = ∞ .

We now state several theorems.

Theorem 1

f(x)  ∞ as x  x_0 ⟺ 1/f(x)  0 as x  x_0 .

Theorem 2

Underscript[lim, xx_0] f(x)/g(x) = 0 ⟹ Underscript[lim, xx_0] g(x)/f(x) = ∞ .

 

The Math Challenge from Last Time

The math challenge was to prove three theorems.  The first of these is, f(x)  L ∧ g(x) M as xx_0  ⟹ f(x) ± g(x) L ± M as xx_0 .  By definition, if we have some ϵ > 0 we can then choose both δ_1 and δ_2 such that

| f(x) - L | <ϵ/2 if 0 < | x - x_0 | < δ_ (1,)

and

| g(x) - M | <ϵ/2 if 0 < | x - x_0 | < δ_2 .

We can justify this since f(x)  L and g(x) M as x - x_0.  We can combine these using addition (or subtraction)

| f(x) ± g(x) - (L ± M) | ≤ | f(x) - L | + | g(x) - M | <ϵ/2 + ϵ/2 = ϵ

so long as 0 < | x - x_0 | < min {δ_1, δ_2}. In other words f(x) ± g(x) L ± M.
    The second theorem to proven states f(x)  L ∧ g(x) M as xx_0  ⟹ f(x) g(x) L M as xx_0 . According to theorem 2 from "Limits By Themselves" we have,

f(x) = L + α(x)

and

g(x) = M + β(x)

where

α(x) 0, β(x) 0, as xx_0 .

So we now have

f(x) g(x) = L M + [b α(x) + a β(x) + α(x) β(x)] .

By applying theorem 3 from "The Nitty Gritty" each term in brackets goes to 0 as xx_0.  If we apply theorem 1 of "Combining Limits" to each pair of terms in brackets we see that the entire quantity in brackets goes to 0.  Therefore, we have f(x) g(x) = L M.

The final theorem stated, f (x)  L ∧ g(x) M as xx_0  ⟹ f(x)/g(x) L/M as xx_0 .  If we write,

f(x)/g(x) - L/M,

this can be cross multiplied,

1/(M g(x)) [M f(x) - L g(x)] .

By applying theorem 2 from "Combining Limits" we see that M g(x)  M^2 as xx_0.  Since

M^2/2<M g(x) < (3 M^2)/2

for a sufficiently small deleted neighborhood of x_0.  However, we have

0<1/(M g(x)) <2/M^2 .

From this we can say that 1/M g(x) is bounded in a deleted neighborhood of x_0.  By theorems 1 and 2 of "Combining Limits" we have

M f(x) - L g(x) ⟹ M L - L M = 0 as xx_0 .

Thus, by theorem 3 of "The Nitty Gritty"

1/(M g(x)) [M f(x) - L g(x)] 0

as xx_0.  Hence,

f(x)/g(x) - L/M  0

as xx_0.  By applying theorem 2 from "Limits By Themselves" we see that

f(x)/g(x)  L/M

as xx_0.

The Math Challenge

Can you prove theorems 1 and 2 of this column?

Math Resources for Limits

Online:

http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node37.html

This is a rather uninspired page, but it is the best one that I could find.

Books:

Richard A. Silverman, 1969, Modern Calculus and Analytic Geometry, MacMillan Company, New York (Dover Publications has reprinted this book with corrections in 2002). This has a very nice chapter on limits that includes a very detailed discussion of infinite limits.

Created by Mathematica  (August 7, 2003)