The Math Forum

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

0 Raised to a Power

Date: 04/25/2002 at 19:38:36
From: Kavi
Subject: 0 raised to a power

My calculus teacher posed a question today. He wanted to know what the 
limit of 0^n as n approaches infinity is. I don't know whether it is 
actually zero or if it is a trick question. Can you explain why it 
equal zero or why it can't equal zero, or if it's undefined?


Date: 04/25/2002 at 21:06:15
From: Doctor Peterson
Subject: Re: 0 raised to a power

Hi, Kavi.

When a mathematician gets a question like this, he doesn't just go 
ask another mathematician; he checks the definitions and finds out 
for himself. That's what definitions are for!

So take the definition of a limit, and see if it says that the limit 
of 0^n is 0. If you have trouble, write back and show me what you've 
found, and I'll help you through it.

- Doctor Peterson, The Math Forum 

Date: 04/25/2002 at 23:20:06
From: Kavi
Subject: 0 raised to a power

Hello again,

After reading your response to my email regarding taking the limit of 
zero to the n power where n approaches infinity, I did a bit of 
reseach on it. I checked in my calculus book on the definition of 
limits and how to calculate them to infinity. I also checked online on 
a precise definition of limits and what occurs if you had to take the 
limit of zero. 

After doing a bit of math on my own, I still can't figure it out. 
Instead of taking the limit of zero to the n power, I decided to see 
how the limit would be if I changed zero to any integer. What I got 
was the limit as n approaches infinity of x^n. With my understanding 
of the definition of limits and my math knowledge, I came to the 
conclusion that this limit is indeterminate. It is indeterminate 
because if x was a negative number raised to n, then the limit would 
switch back and forth between negative and positive values. If x was 
positive, then the limit of x^n would be positive infinity. If I 
assume, (I don't know if it is right to assume this or not) that zero 
will also follow this idea, then zero to the nth power would also be 
indeterminate. I don't know if this is correct but it seems logical to 
me. Is my logic accurate, or have i missed something if figuring this 
limit out.


Date: 04/26/2002 at 08:46:13
From: Doctor Peterson
Subject: Re: 0 raised to a power

Hi, Kavi.

But you were asked to find the limit of 0^n, not x^n, so you're really 
solving the wrong problem. Let's back up and do the original problem 

I suggested you see whether the definition of a limit supported saying 
that lim[n->oo] 0^n = 0. Here's the definition:

    lim f(n) = L


    for any e>0 there is an N such that |f(x)-L|<e whenever n>N

Applying that to this case,

    lim 0^n = 0


    for any e>0 there is an N such that |0^n|<e whenever n>N

But for any (finite) n, 0^n is zero, so |0^n| is ALWAYS less than any 
e, no matter what N you choose. So, yes, the limit is zero.

Now, if you look at x^n, you have to consider what is changing. You 
are still talking about only taking n to infinity, so x is fixed. And 
for any fixed x>1, x^n gets larger and larger as n increases, so the
limit is infinite. If x=1, x^n is always 1, so the limit is 1. For a 
fixed x with -1<x<1, x^n gets closer to 0 as n increases, so the 
limit is 0. Finally, for x<=-1, you are right that |x^n| gets larger
(except when x=-1), but x^n alternates sign, so there is no limit. 
That means we have four different cases depending on the value of x. 
This doesn't make your limit indeterminate. In fact, because x^n 
approaches zero for x near zero (despite the fact that it alternates 
sign for negative x), even the expression 0^infinity is  not 
indeterminate, since the limit of x^y, as x and y approach 0 and 
infinity independently, is zero. But in any case, broadening the 
problem didn't help in solving the original problem.

This is why I sent you back to the definition, rather than just 
suggesting you think about it. Definitions are what math is all about, 
and in this case the definition confirms your original impression. So 
don't be afraid just to take the obvious guess and check it out, 
before looking for other possibilities.

- Doctor Peterson, The Math Forum 

Date: 04/26/2002 at 20:22:28
From: Kavi
Subject: 0 raised to a power

Dr. Peterson,

Thank you for sending me the definition of the limit to help me 
understand the problem. That really helped me to understand why the 
limit of 0^nth power and n approaches infinity is zero. I also 
appreciate you explaining why expanding the problem to include all 
integers didn't help solve the problem. Your explanation has broadened 
my understanding of limits even though I had learned them well over a 
year ago.

Thank you,

Date: 04/26/2002 at 22:19:04
From: Doctor Peterson
Subject: Re: 0 raised to a power

Hi, Kavi.

Thanks for writing back. There's a lot of subtlety in limits; can you 
imagine the confusion for the couple of hundred years between the 
invention of calculus and the introduction of a careful definition of 
the limit?

- Doctor Peterson, The Math Forum 
Associated Topics:
College Analysis
College Calculus
College Definitions
High School Analysis
High School Calculus
High School Definitions

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- The Math Forum at NCTM. All rights reserved.