Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

The limit of the nth root of n as n approaches infinity equals.... 1!!! I don't know about you guys but I find this extremely remarkable!

The proof of this uses just a simple common technique. I'll use the method of finding the limit of ln (n^(1/n)) as n approaches infinity, and raising e to the power of that limit.

limit of ln (n^(1/n)) as n approaches infinity

= limit of 1/n ln( n) as n approaches infinity = limit of (ln n)/n as n approaches infinity ( = ∞/∞) since this is an indeterminant form, we use L' hopitals rule, 1/n / 1 = 1/n. As n approaches infinity 1/n = 0.

e^0 = 1 so the limit of the the nth root of n as n approaches infinity is one! Thats just awsome! Check it on a calculator if you don't believe me.

*Last edited by mikau (2006-02-24 11:06:24)*

A logarithm is just a misspelled algorithm.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

This reminds me of e in some ways. This seems to suggest 1^∞ = ∞ but this is impossible. Therefore it must be something like the limit of (1 + 1/x )^n as n approaches infinity and x is some function of n, and n does not equal x. (otherwise this would be e). I wonder if we could figure out the relation between n and x...

A logarithm is just a misspelled algorithm.

Offline

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

Given that 1/∞ = 0, it's not that surprising at all...

El que pega primero pega dos veces.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Yes and no. I mean, if you have 1/n, when n becomes significantly large, the exponant tends to be very small, but the base tends to get very large. Just like the definition of e. limit of (1 + 1/x)^x as x approaches infinity. You're inclined to say 1/x = 0 and eliminate it, but the infinitely small difference, to the infinitieth power makes a difference. Likewise, I was inclined to believe the smallness of the exponant would be "canceled out" by the hugeness of the base.

A logarithm is just a misspelled algorithm.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I haven't tried to understand your explanations, but it isn't remarkable to me.

#*#=2

#*#*#=3

#*#*#*#=4

#*#*#*#*#=5

#*#*#*#*#*#=6

The #'s go up slowly, but multiplying is so powerful, why wouldn't you expect to

approach one from above? I would.

'Course I can't remember L'Hopital's Rule by its name. I have a lot to review.

Don't explain for me, I'll do some research...

**igloo** **myrtilles** **fourmis**

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

maybe you guys are more familiar with limits then me and it doesn't come to such a suprise.

At any rate, the reason I didn't conclude that from the start, I could have looked at the exponant 1/n and figured this comes to zero so its one no matter what, but singling out and isolating a piece of a limit and evaluating can be dangerous, ESPECIALLY when they are directly connected to infinitly large numbers. I repeat: In the definition of e: Limit of (1 + 1/n)^n as n approaches infinity. Your inclined to notice the 1/n term and rewrite it as zero, but this infinitly small fraction is raised to an infinitly large power! So it does make a difference! Here we have the reverse, an infinitly large number being raised to an infinitly small power. For all I knew it might have ended up being 1/e or something like that. Thats why I made no assumptions.

Like I said, when infinitly small and infinitly large come in contact, sometimes they "cancel eachother out" so isolating each individual limit can be dangerous. Don't you agree?

*Last edited by mikau (2006-02-24 17:40:58)*

A logarithm is just a misspelled algorithm.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

.99 ^ big number, goes down.

1.01^ big number, goes up.

**igloo** **myrtilles** **fourmis**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm not really suprised at 1, but I really like the way mikau took the limit. Never would have thought of doing that.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

you mean taking the natural log, finding the limit, and raising e to that power? Thats a pretty common technique in my mathbook.

Anyways, the title of this thread is "who woulda thunk it?" clearly the answer is "everyone woulda thunk it.... except mikau..."

I feel stupid...

*Last edited by mikau (2006-02-25 08:29:18)*

A logarithm is just a misspelled algorithm.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Wait a minute...

It was shown somewhere else that this equation:

Had the solution of x = the nth root of n.

Surely, what you've just said means that:

Why did the vector cross the road?

It wanted to be normal.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

now I'm confused.

technically the limit I found seems to suggest 1^∞ = ∞ but this is obviously not true. This is just the limit. 1 to any power = 1 no matter what. The nth root of n as n approaches infinity, tends to be 1, but 1 is the limit, it can never equal 1.

A logarithm is just a misspelled algorithm.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,582

And, as John said, anything below 1 goes to 0.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

Pages: **1**