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### 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?

Thanks,
Kavi
```

```
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
http://mathforum.org/dr.math/
```

```
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.

Thanks
Kavi
```

```
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
first.

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
n->oo

if

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
n->oo

if

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
http://mathforum.org/dr.math/
```

```
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,
Kavi
```

```
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
http://mathforum.org/dr.math/
```
Associated Topics:
College Analysis
College Calculus
College Definitions
High School Analysis
High School Calculus
High School Definitions

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