Orders of Infinity

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Special Mathematica Functions used in this NoteBook

Needs["Graphics`Colors`"]

Notebook Overview

This NoteBook compares the rate at which various functions tend to infinity.  Comparison of different powers is easy.  An important new fact of Chapter 8 is that Logs grow slower than any power and exponentials grow faster than any power.

Intuitive measures of growth can be seen in the NoteBooks ExpGth and LogGth from Chapte 28 on High School Review.

The first step analytically is to learn that finding the eventual size of the ratio

g[x]/f[x]

for two functions that already tend to Infinity is a better measure of their rates of growth at Infinity than their actual size at a particular  x, or even of the eventual size of the difference between them.  Once you see this, the next theorem says, "Exponentials always give a higher order of infinity than powers." and "Powers, even roots, always give a higher order of infinity than Log."

Theorem on Exponential and Logarithmic Orders of Infinity

Let  p > 0  be any positive real number.  Then

Limit[((E^t)/(t^p)), t -> ∞] = ∞

and

Limit[((t^p)/Log)[t], t -> ∞] = ∞

References to the Text

This NoteBook goes with Section 8.3 of the text.

"Orders of Infinity" as Ratio Limits

A Basic Fact

Limits of positive power functions tend to infinity, even for small positive powers,

(Underscript[Lim, x -> ∞]) (x^p) = +∞,  for  p > 0,

but the smaller the power, the slower they grow.  Here's a first look:

Clear[f, x, p]; f[x_] := (x^p); p = 3.1Limit[f[x], x -> ∞]\(plot1\  = \ Plot[f[x], {x, 0, 1000}, PlotStyle -> {Red}]; 
p = 1.3\)Limit[f[x], x -> ∞]\(plot2\  = \ Plot[f[x], {x, 0, 1000}, PlotStyle -> {Blue}]; 

    p = 0.13\)Limit[f[x], x -> ∞]\(plot3\  = \ Plot[f[x], {x, 0, 1000}, PlotStyle -> {DarkGreen}]; 

    Show[{plot1, plot2, plot3}]; \)

Practice Exercises

What is the limit as  x ->∞ of negative power functions?

(Underscript[Lim, x -> ∞]) (x^q) = ??,  for  q < 0?

Illustrate this limit for some negative powers using machine computations as above.

Example 8.2: Eventual Size is What Counts

Consider the functions  f[x] = 500 x  and  g[x] = ((x^2)/500).  The first picture below is WRONG in the sense that it does not describe the eventual behavior.  Once  x  is "big enough" the square with a tiny coefficient overpowers the linear term with the huge coefficient.  The larger power will always win ... eventually.  We want a way to formulate this idea analytically.

Clear[f, x, g, p, q]; h[x_] := 500 x; j[x_] := ((x^2)/500); Plot[{h[x], j[x]}, {x, 0, 100}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}];

n = 100; Do[n = 10 n; 
      Plot[{h[x], j[x]}, {x, 0, n}, 
          PlotStyle 
            -> {{Blue, AbsoluteThickness[1]}, {Red, 
                AbsoluteThickness[3]}}];, {k, 1, 4}]

Simple Power Comparisons

We want to formualte the idea that bigger positive powers go to infinity faster than small ones.  We might graph both functions as follows:

Clear[f, x, g, p, q]; f[x_] := (x^p); g[x_] := (x^q); p = 3; q = 4; Plot[{f[x], g[x]}, {x, 0, 100}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}];

Two functions can tend to Infinity at comparable rates and yet still have one beat the other.  For example, both of the linear functions below tend to Infinity, and  5 x tends faster in the sense that  3 x - 2 x even tends to infinity:

(Underscript[Lim, x -> ∞]) (5 x  -  2 x)  =  
        (Underscript[Lim, x -> ∞]) 3  x  =  ∞

Clear[f, x, g]; f[x_] := 2 x; g[x_] := 5 x; Limit[g[x] - f[x], x -> ∞]Plot[{f[x], g[x]}, {x, 0, 100}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}]; 
    Plot[g[x] - f[x], {x, 0, 100}, 
      PlotStyle -> {{DarkGreen, AbsoluteThickness[2]}}];

Comparison by a difference is not the strongest one we can make.  Consider the functions  h[x] = 500 x  and  j[x] = x^2/500 again.  The difference between these functions tends to Infinity, but a stronger comparison is given by the RATIO.  Any to linear functions with positive coefficients tend to Infinity and their ratio satisfies

(Underscript[Lim, x -> ∞]) (((a x )/(b x)))  =  
        (a/b), a constant.

In fact any two functions with the same power satisfy

(Underscript[Lim, x -> ∞]) (((a (x^p))/(b (x^p))))  =  
        (a/b), a constant.

The ratio comparison gives us an analytical way to say that eventually  g[x]  gets MUCH bigger than  f[x]  when

(Underscript[Lim, x -> ∞]) (g[x]/f[x])= ∞

Analytically this isn't hard,

(Underscript[Lim, x -> ∞]) (((a (x^p))/(b (x^p)))) 
       = (b/a) (Underscript[Lim, x -> ∞]) (x^(q - p))

so if  q > p, q-p > 0 and the limit is Infinity no matter what the positive values of  a and  b.

Clear[f, x, g, h, j]; f[x_] := 2 x; g[x_] := 5 x; h[x_] := 500 x; j[x_] := ((x^2)/500); Limit[g[x] - f[x], x -> ∞]Limit[j[x] - h[x], x -> ∞]Limit[g[x]/f[x], x -> ∞]Limit[j[x]/h[x], x -> ∞]Plot[Evaluate[{g[x]/f[x], j[x]/h[x]}], {x, 1, (10^6)}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}];

Practice Exercise

Suppose  f[x]  and  g[x]  are both functions that tend to infinity as  x  tends to Infinity.  
	
1)  Explain in your own words why the
	
	Lim g[x]/f[x]  = Infinity
	
as  x -> Infinity, means that  g[x]  is a larger "order of infinity" than  f[x], as  x -> Infinity.  

2)  Also explain in your own words why the
	
	Lim f[x]/g[x]  = 0
	
as  x -> Infinity, means the same thing as the previous limit result.

3)  Suppose 	Lim g[x]/f[x]  = Infinity, but we don't know that both f[x] and g[x] tend to infinity.  Does the ratio limit mean that  g[x]  tends to infinity faster than  f[x]?

Ratios of Combinations of Powers

The limit of the ratio of sums of functions depends on the "leading" term.  When one term in the numerator grows fastest, the limit is still Infinity.  For example,

(Underscript[Lim, x -> ∞]) 
        (((2 (x^3) - 4 (x^2) + 5 x - 7)/(800 (x^2) + 900 x + 1000)))
           =(Underscript[Lim, x -> ∞]) ((2 x - 4 + (5/x) - (7/(x^2)))/(800 + (900/x) + (1000/(x^2))))
 = (1/400) (Underscript[Lim, x -> ∞]) 
        ((x - (1/200) + (5/(800 x)) - (7/(800 (x^2))))/(1 + 
            (900/(800 x)) + (1000/(800 (x^2))))) =(((BIG - (1/200) + tiny))/(400 (1 + tiny + teeny)))= BIG

Clear[f, g, x]; f[x_] := 800 (x^2) + 900 x + 1000; g[x_] := 2 (x^3) - 4 (x^2) + 5 x - 7; Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

Log and Exponential Orders of Infinity

Powers and Exponentials

Try these limits:

Clear[x, f, g]; f[x_] := (x^3); g[x_] := (2^x); Limit[g[x]/f[x], x -> ∞]Plot[{f[x], g[x]}, {x, 1, 10}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}]; 
    Plot[g[x]/f[x], {x, 1, 10}, 
      PlotStyle -> {{Green, AbsoluteThickness[2]}}]; Plot[{f[x], g[x]}, {x, 1, 20}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}]; 
    Plot[g[x]/f[x], {x, 1, 20}, 
      PlotStyle -> {{Green, AbsoluteThickness[2]}}];

Questions

In the previous output the thick red curve is supposed to be above the thin blue one.  However, for  1<x<10, this is not true.  When does g[x]  rise above  f[x]?

Practice Exercise

Compare the growth of  f[x] = (x^5)  and  g[x] = (1.5^x)  as  x -> Infinity and show analytically and graphicaly that  g[x]  is eventually larger.  Where does  g[x]  first exceed  f[x]?

Powers and Logs

Try these limits:

Clear[x, f, g]; f[x_] := Log[x]; g[x_] := (x^(1/10)); Limit[g[x]/f[x], x -> ∞]Plot[{f[x], g[x]}, {x, 2, 10}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}]; 
    \(Plot[g[x]\/f[x], {x, 2, 10}, 
      PlotStyle -> {{DarkGreen, AbsoluteThickness[2]}}]; 

    \(Plot[{f[x], g[x]}, {x, 2, 10\^20}, 
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}]; 
    \)\)Plot[g[x]/f[x], {x, 2, (10^20)}, 
      PlotStyle -> {{DarkGreen, AbsoluteThickness[2]}}];

Questions

In the previous output the thick red curve is supposed to be above the thin blue one.  However, for  2<x<10, this is not true.  When does g[x]  rise above  f[x]?

Practice Exercises

Compare the growth of  f[x] = Log[x]  and  g[x] = Sqrt[x]  as  x -> Infinity and show analytically and graphicaly that  g[x]  is eventually larger.  Where does  g[x]  first exceed  f[x]?

Indeterminate Forms as Orders of Infinity

What's Determined and What's Not?

The Limit of   ((2^(-x))/(x^3))   is trivial, because BOTH functions are headed the same way.  Re-write the expression as ((2^(-x))/(x^3)) = (1/((2^x) (x^3)))  so both terms in the denominator tend to infinity:

Limit[(1/((x^3) (2^x))), x -> ∞]

Converting Limits at Zero to Orders of Infinity

Mathematica can compute the following limits, but they can also be put in the form of orders of infinity bu changes of variable.

Limit[(x^x), x -> 0]

Limit[x Log[x], x -> 0]

The function (x^x)  can be re-written  x = Exp[ Log[x] ], reducing th first limit to the second.

The limit as  x -> 0 of  x Log[x]  can be changed into a limit toward infinity by letting  u = 1/x, so

x Log[x] = (1/u) Log[1/u] = (1/u) Log[ u^(-1) ] = -Log[u]/u

and  u -> Infinity as  x -> 0.

The limit

Lim Log[u]/u

as  u -> Infinity is  0  by the reciprocal version of the orders of infinity theorem.

Exercises

Exercise 8.3.1: Ratios of Powers

Show that each of the functions  g[x]  below tends to infinity faster than the corresponding  f[x]:

a)  g[x] = (x^5)  &   f[x] = (x^3)

b)  g[x] = (x^(2  ))&   f[x] = (Sqrt[x])

c)  g[x] = (Sqrt[x])  &   f[x] = (x^(1/3))

d) g[x] = ((x^5)/(10  ))  &  f[x] = 10 (x^3)

e) g[x] = ((x^2)/2)  &  f[x] = 2 (Sqrt[x])

f)  g[x] = ((Sqrt[x])/2)  &  f[x] = 3 (x^(1/3))

Exercise 8.3.2: Ratios of Powers and Orders of Infinity

Show why the "order of infinity" comparing the eventual size of expressions like

f[x] = a1 (x^p1) + b1 (x^q1) + c1 (x^r1)
and		
	g[x] = a2 (x^p2) + b (x^q2) + c2 (x^r2)

for positive constants  a1, a2, b1, b2, c1, c2, p1, p2, q1, q2, r1, r2, depends only on the largest power.  For example, if p2 is greatest, then  g[x]  is eventually largest because ...??

What happens if p1 = p2 and this is larger than all the other terms?  What is the limit of the ratio  g[x]/f[x]?  Which one is eventually largest (in terms of the other parameters?)

Exercise 8.3.3: Log and Exp Limits

Compute the following limits:

a)  Limit[(x^p) Log[x], x -> 0],  p>0

b)  Limit[(x^(1/x)), x -> ∞]

c)  Limit[(d^((1/Log)[d])), d -> 0, d > 0]

d)  Limit[(x^p) Log[x], x -> ∞],  p>0

e)  Limit[((1 + k d)/d), d -> 0],  k > 0

f)  Limit[(d^(((d + k)/Log)[d])), d -> 0, d > 0], k > 0

Exercise 8.3.4: Different Exponentials

Suppose that  a  and  b  satisfy  b > a > 1.  Find

Limit[((b^x)/(a^x)), x -> ∞]

For example,
	b = 2
	a = 3/2
	Limit[ b^x/a^x , x -> Infinity ] = ?

Exercise 8.3.5: Negative? Exponentials

If the base  b  of an exponential function  f[x] = b^x  satisfies  b < 1, we can think of this as a "negative" exponential in the following sense.

a)  Solve for a constant  k  so that  b^x = Exp[ k x ],  k = ?

b)  Show that  0<b<1 makes  k < 0

c)  Show that when  b < 1, Limit[ b^x, x->Infinity ] = 0