Can the Squeeze Theorem Determine the Limit of n^n/n! as x Approaches 0?

[Ignea_unda]

[SOLVED] Factorial Limits

Homework Statement

lim n^n
x->00 n!

Homework Equations

Instructor said to use the Squeeze theorem.

The Attempt at a Solution

So far I have not been able to come up with much. I have looked at breaking the top apart into (n)(n)(n)...(n) and the bottom into n(n-1)(n-2)...(2)(1). My instinct when plugging numbers in for examples says that it should diverge, but when I apply Induction to prove this hypothesis, I get a contradiction. I know I'm over thinking this one (or at least I hope I am). Thanks for your help

 

[Dick]

Your instincts are right. Write it as (n/n)*(n/(n-1))*(n/(n-2)*...*(n/1). For i>=n/2. n/(n-i)>=2. For i<=n/2, n/(n-i)>=1. So the whole product must be greater than or equal to 2^(n/2), right? You may need to adjust a few details for n odd, and there actually may be a neater comparison, but that's what came mind first.

 

[Ignea_unda]

Thanks for the help. I looked at it again and talked to my professor. I was thinking that you couldn't use the squeeze theorem to prove that the limit was inifinity. Since it's not, I just compared it to n. Since n! >= n for all n, and n diverges to infinity, so does n^n/n!. Once again thanks for the quick reply.

 

[cjs_101]

I have the exact same problem, however it is the inverse:
lim n!
n->00 n^n

Fairly intuitively this limit will be zero, however I need to 'use the squeeze rule'
The lower limit can easily be 1/n^n whose limit is zero. However I cannot think of an upper limit to 'squeeze' my limit between. It must be greater than n!/n^n and yet must also have a limit of zero. Please help

 

[Dick]

Write it as (n/n)*((n-1)/n)*((n-2)/n)*...(1/n). All of the terms in the product are less than or equal to 1 and positive. n/2 terms are less then or equal to 1/2. Does that suggest a squeeze strategy?

 

[cjs_101]

This is exactly what I was thinking, but won't this only show that the limit is (considerably) less than 1/2? I don't believe this proves it is zero.

 

[Dick]

It shows that its less than (1/2)^(n/2). That is 'considerably' less than 1/2. What happens as n goes to infinity?

 

[cjs_101]

thanks, i overlooked that, that solves my problem