Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Limits Approaching Infinity


Date: 02/16/2002 at 12:34:58
From: Miranda Cir
Subject: Limits approaching infinity

I am lost when it comes to limits approaching infinity. Is the only 
way to answer them by memorizing all the general questions you may 
come across?

ex:  lim --> infinity (1/2)^n = 0
 and lim --> infinity 4^(-infinity) = infinity
 and lim --> infinity 4^[1/(-infin)] = 1

I'm confused. Can you give me some advice?


Date: 02/16/2002 at 23:09:00
From: Doctor Peterson
Subject: Re: Limits approaching infinity

Hi, Miranda.

First, we have to correct what you wrote. A limit has to state what 
variable is going to infinity. In the first, it is presumably n; but 
the others have no variable. I think you meant

    lim[n->inf] (1/2)^n = 0

    lim[n->inf] 4^-n = inf

    lim[n->inf] 4^(1/-n) = 1

There are only a couple of basic facts you should learn; one is:

    lim[n->inf] 1/n = 0

This says that if you divide by larger and larger numbers, the 
quotient goes to zero. Memorize the shape of the graph of this 
reciprocal function; it's extremely useful and common.

Let's take your examples one at a time.

    lim[n->inf] (1/2)^n = lim[n->inf] 1/(2^n) = lim[m->inf] 1/m = 0

I first simplified the exponent using exponent properties so I had the 
reciprocal of something. Then I noticed that 2^n goes to infinity when 
n goes to infinity; by calling that m, I have the limit of 1/m, which 
is 0 according to the fact above.

A general fact worth knowing is that a^n goes to zero if a<1, and to 
infinity if a>1. You might want to graph this for several values of a 
to see what it means. Memorize the shape of the graph in these two 
cases. (And see what happens if a=1, too.)

Your second problem:

    lim[n->inf] 4^-n = lim[n->inf] 1/(4^n) = 0

by the same reasoning; it's just written differently at first. Did I 
correct the problem correctly?

Finally, 

lim[n->inf] 4^(1/-n) = lim[n->inf] 4^(-1/n) = lim[p->0]4^-p = 4^-0 = 1

This time I looked at the exponent -1/n and saw that it goes to zero; 
since 4^0 is 1, we don't need any tricks to find the limit.

It looks as if a general principle is: look at the base and the 
exponent, and see where they are going. As I said before, if the 
exponent is going to infinity, the limit depends on the base. But if 
the exponent is going to zero, the limit is 1 unless the base is, or 
is going to, 0, which would be indeterminate.

If you need more help, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 02/17/2002 at 12:33:25
From: Miranda
Subject: Limits approaching infinity

I just wanted to thank you! This website is such an amazing help, I am 
definitely going to be coming back for more questions! 

Thanks again.
    
Associated Topics:
High School Calculus
High School Exponents

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/