0 Raised to a PowerDate: 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/ |
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