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        riddles >> medium >> Evaluate limit
        
(Message started by: Benny on Mar 20^th, 2012, 11:32am)

Title: Evaluate limit
Post by Benny on Mar 20^th, 2012, 11:32am

What is 1^n, with n --> infinity ?

I saw different versions of this limit. It's either indeterminate, or if I write (1^n) for n=inf in Wolfram Alpha or Mathematica, the result is 1.

Or we could write 1^infinity = 1*1*1.....*1 = 1

the final answer could be any
number, such as 1, or infinity, or undefined.

Why couldn't we say and settle for:

If n --> infinity Limit of 1^n is equivalent to asking what is the limit of

x --> 0 limit of (x+1)^1/x = e

So, n --> infinity limit of 1^n = e

Title: Re: Evaluate limit
Post by towr on Mar 20^th, 2012, 12:03pm

Err, no..
You can't take limits in separate parts, so the expression limit x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/downarrow.gif0 (x+1)^1/x = limit n->inf (1/n+1)ⁿ, which equals e, is not the same expression as limit n->inf (limit n->inf (1/n+1))ⁿ = limit n->inf 1ⁿ, which equals 1.

Limits have a rigorous definition, so I suggest using those, rather than making up rules yourself. (Which isn't to say I know all the fine points myself, but when you start getting silly results it's a sure sign you should get out the text-book and try doing it the right way first.)

For example, the epsilon-delta definition of limits says:
A function f(z) is said to have a limit lim_(z->a) f(z)=c if, for all epsilon>0, there exists a delta>0 such that |f(z)-c|<epsilon whenever 0<|z-a|<delta.
So, for f(z) = 1^z, we have for all delta that |f(z)-1| = 0 which is smaller than any positive epsilon. Therefor c=1.

Title: Re: Evaluate limit
Post by Benny on Mar 20^th, 2012, 12:12pm

(1)
Let me ask you this: Isn't it true that sometimes when it is difficult to evaluate a limit we use equivalent expressions?

(2) Why is limit n -> inf 1^n = 1 ? How can we prove it?

Title: Re: Evaluate limit
Post by pex on Mar 20^th, 2012, 12:58pm

on 03/20/12 at 12:12:36, Benny wrote:

(1) Let me ask you this: Isn't it true that sometimes when it is difficult to evaluate a limit we use equivalent expressions?

Yes. But they have to be... well, equivalent. Yours aren't.

on 03/20/12 at 12:12:36, Benny wrote:

(2) Why is limit n -> inf 1^n = 1 ? How can we prove it?

on 03/20/12 at 12:03:17, towr wrote:

For example, the epsilon-delta definition of limits says:
A function f(z) is said to have a limit lim_(z->a) f(z)=c if, for all epsilon>0, there exists a delta>0 such that |f(z)-c|<epsilon whenever 0<|z-a|<delta.
So, for f(z) = 1^z, we have for all delta that |f(z)-1| = 0 which is smaller than any positive epsilon. Therefor c=1.

Title: Re: Evaluate limit
Post by Benny on Mar 20^th, 2012, 1:45pm

Thank you, pex, for your patience.

In Wolfram Alpha

http://www.wolframalpha.com/input/?i=1%5Einfinity

Result: indeterminate

and

http://www.wolframalpha.com/input/?i=%281%5En%29+for+n%3Dinf

Result: 1

Title: Re: Evaluate limit
Post by towr on Mar 20^th, 2012, 1:57pm

The 'problem' there is that 1^inf isn't an equivalent expression to lim n->inf 1ⁿ.

For example take the function f(x) = 1 if (x>0)
lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/downarrow.gif1 f(x) = 1, but f(0) is indeterminate, since we have only assigned values for x > 0.
The limit of a function at a point (approached from some direction) isn't necessarily the same as the value of that function at that point; it may not even have a value at that point (as in the example I just gave). However, the function converges on a value as we approach that point, and that value it converges on is the limit value.

It may sometimes be convenient to treat infinity as a number, but it isn't one in regular mathematics. So 1^inf isn't actually a valid expression, since powers are only defined for numbers. Therefore 1^inf has the same value as 1^pancake (albeit not according to wolframalpha).
In limits and integrals when we use infinity we actually avoid that, because we sort of look only at all number up to, but not including, infinity.

Title: Re: Evaluate limit
Post by Benny on Mar 20^th, 2012, 2:13pm

Thank you Towr.

You're right. I see now the 2 expressions in WolframAlpha are not equivalent.

Infinity should not be treated as a number, and
1^n as n->oo is indeed 1 but in the expression 1^inf since infinity is not a number

Can we say 1 to power infinity is undefined?

Title: Re: Evaluate limit
Post by pex on Mar 20^th, 2012, 2:26pm

on 03/20/12 at 14:13:08, Benny wrote:

Can we say 1 to power infinity is undefined?

That's correct - "a to the power b" is not defined if a and b are not both numbers. "1 to the power infinity" is about as meaningful as "1 to the power pancake", to take towr's example.

Sorry for my harsh reply earlier tonight. I was just a little put off by you asking for a proof that was provided in the post that you were relplying to...

Title: Re: Evaluate limit
Post by Benny on Mar 20^th, 2012, 2:53pm

on 03/20/12 at 14:26:12, pex wrote:

That's correct - "a to the power b" is not defined if a and b are not both numbers. "1 to the power infinity" is about as meaningful as "1 to the power pancake", to take towr's example.

At first, I thought that 1^infinity must be undefined. Then, I used Wolfram Alpha search engine, which gave me two different versions. Then, after interacting on this forum, I realized that these two expressions are not equivalent... no matter how much we like them to be.

which brings me to the expression (x+1)^1/x
a classic case when we look for something in order to achieve the desired outcome.

Quote:

Sorry for my harsh reply earlier tonight. I was just a little put off by you asking for a proof that was provided in the post that you were relplying to...

No worries. I'm very patient. Mea Culpa. I didn't read carefully the second part of towr's answer.

Thank you pex and thank you towr