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

In summary, the given problem involves finding the limit of n^n/n!, and the squeeze theorem is recommended for solving it. The solution involves breaking down the fraction and using a comparison to show that the limit is equal to zero. Another similar problem is also discussed and solved using the same method.
  • #1
Ignea_unda
133
0
[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
 
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  • #2
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.
 
  • #3
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.
 
  • #4


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
 
  • #5


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?
 
  • #6


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.
 
  • #7


It shows that its less than (1/2)^(n/2). That is 'considerably' less than 1/2. What happens as n goes to infinity?
 
  • #8


thanks, i overlooked that, that solves my problem
 

What is a factorial limit?

A factorial limit is a mathematical concept that refers to the limit of a sequence of numbers in which each term is multiplied by all the previous terms. It is represented by the symbol "!" and is often used in probability and combinatorics.

How do you calculate a factorial limit?

The factorial limit of a number can be calculated by multiplying the number with all the numbers that come before it. For example, the factorial limit of 5 would be 5! = 5 x 4 x 3 x 2 x 1 = 120.

What is the significance of factorial limits in probability?

In probability, factorial limits are used to calculate the number of possible outcomes in an experiment. For example, if you have 5 different colored balls and want to know how many different ways you can choose 3 of them, you would use the permutation formula nPr = n!/(n-r)! = 5!/(5-3)! = 5 x 4 x 3 = 60.

Can factorial limits be negative?

No, factorial limits are only defined for non-negative integers. This is because the concept of factorial limits involves multiplying all the numbers before it, and negative numbers do not have a defined factorial.

What are some real-world applications of factorial limits?

Factorial limits have various real-world applications, such as in genetics to calculate the number of possible gene combinations, in economics to calculate the number of ways a company can distribute its resources, and in computer science to calculate the number of possible paths in a network.

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