How do i prove that the limit of x^n / n is 0?

  • Thread starter burhan1
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In summary, the problem is that x^n/n! -> 0 for x>1 but not for x=1. You need to find a way to prove that x^n/n! -> 0 for all x.
  • #1
burhan1
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Homework Statement


x^n / n! -> 0 for any value of x and n -> 0


Homework Equations





The Attempt at a Solution

 
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  • #2
What have you tried to do so far?
 
  • #3
I know generally how to do it i just don't get how to write it in a formal proof. like x^n is always going to be approaching infinity slower than (n!) for x > 1 and for x = 1 its just 1/n! but what about x =< -1? because this doesn't converge to any number. however (n!) will always be greater than |x| soo...

hey wait maybe i could prove that |x^n|/(n!) -> 0 and use the absolute value rule...
 
  • #4
burhan1 said:

Homework Statement


x^n / n! -> 0 for any value of x and n -> 0
I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?
 
  • #5
Mark44 said:
I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?

I think so too. The limit of xn/n! as n--->0 is 1, not 0 as x0=1 and 0!=1.

As for xn/n! as n--->∞, try expanding the numerator and denominator of the limit. What does xn look like? What does n! look like?
 
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  • #6
Oops yeah n -> infinity. sorry about that =/
 
  • #7
well x^n looks like a hell lotta x's multiplied together and n! looks like n(n-1)(n-2)...! and i realize that for any value of x, there exists n s.t n is wayyyyyy greater than x so n! will be wayyy greater than x^n but how do i prove itt... :S
 
  • #8
It's impossible to tell you how to prove this without knowing what you have available! In my oppinion, the simplest way is to note that the series, [itex]\sum x^n/n![/itex] converges to [itex]e^x[/itex]. Since it converges, the sequence of terms, [itex]x^n/n![/itex] must converge to 0. But you may not be able to use that.
 
  • #9
im a first year undergraduate i have nothing on me panicking like crazy xD

and that didnt help TT_TT
 
  • #10
Prove that for n large
(2/n)n<1/n!<(3/n)n
 
  • #11
Show that [x^(n+1)/(n+1)!] < [x^n/n!] when n>>x.
Then [x^(n+1)/(n+1)!] / [x^n/n!] = x/(n+1) = 0 when n→∞.
 
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  • #12
ive already tried those but we don't have sufficient knowledge to use that stuff yet.

heres what I am trying:
Prove that a_n = (x^n/n!) -> 0 for all x

Choose ε > 0. Then there exists for all N ,which belongs to natural numbers, n>N and |a_n| < ε
so, |(x^n/n!)| < ε => |x^n| < n!*ε (since n! is positive the absolute sign is not significant?)
so |x|/[ε^(1/n)] < [n!^(1/n)].

so choose any natural number N s.t. N > |x|/[ε^(1/n)]

If n>N then x^n/n! < x^N/N! < ε

which means a_n -> 0.

is this correct in any sense? =S
 
  • #13
Can you prove that
lim (a x/n)n=0?
because
(2/n)n<1/n!<(3/n)n
implies
xn/n!~(a x/n)n
 
  • #14
no that is the next question lurf :P

i've done it guys. thanks anyway for the help!
 

1. How do I prove that the limit of x^n / n is 0?

The limit of x^n / n can be proven to be 0 using the squeeze theorem. This theorem states that if a function is sandwiched between two other functions and the limit of those two functions is 0, then the limit of the middle function must also be 0. In this case, we can choose the functions x^n and 1/n as the upper and lower bounds, respectively, for x^n / n. Since the limit of x^n is 0 and the limit of 1/n is also 0, the limit of x^n / n must be 0 as well.

2. Can I use L'Hopital's rule to prove the limit of x^n / n is 0?

Yes, L'Hopital's rule can also be used to prove that the limit of x^n / n is 0. This rule states that if the limit of two functions is indeterminate (such as 0/0), then the limit of their derivatives will be equal to the limit of the original functions. In this case, the derivative of x^n is nx^(n-1) and the derivative of n is 1. So, the limit of x^n / n can be rewritten as the limit of nx^(n-1) / 1, which is equal to 0 as n approaches infinity.

3. Is there a geometric interpretation of the limit of x^n / n?

Yes, there is a geometric interpretation of the limit of x^n / n. This limit represents the slope of the tangent line to the graph of y = x^n at x = 0. As n approaches infinity, the graph of y = x^n becomes steeper and steeper at x = 0, resulting in a slope of 0 at this point.

4. Can I use the binomial theorem to prove the limit of x^n / n is 0?

Yes, the binomial theorem can also be used to prove the limit of x^n / n is 0. The binomial theorem states that (1+x)^n = 1 + nx + (n(n-1)/2!)x^2 + ... + x^n. By substituting x = 1/n, we get (1+1/n)^n = 1 + 1 + (1/2!)(1/n)^2 + ... + (1/n)^n. As n approaches infinity, all terms except for 1/n^n become insignificant, resulting in a limit of 1/n^n. This is equivalent to the limit of x^n / n, which is also equal to 0 as n approaches infinity.

5. Why is it important to prove the limit of x^n / n is 0?

Proving the limit of x^n / n is 0 is important because it is a fundamental result in calculus and is used in many other proofs and applications. It also helps to understand the behavior of functions as they approach infinity and provides a basis for understanding more complex limits. This limit is also commonly used in the study of series and sequences, making it an important concept for many mathematical and scientific disciplines.

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