# Confused by a simple limit

1. Dec 7, 2011

### hamsterman

What is the value of
$\lim\limits_{n\rightarrow\infty}\sup \left\{\frac{n}{x^n}:x\in\left( 1; \infty\right)\right\}$
It seems to be 0, but what if $x = 1+\frac{1}{n}$? In that case $x^n = e$ and the above limit is then $+\infty$, isn't it? I have a feeling I'm somehow wrong, but if I'm not, for what x is the above limit equal to 0 ?

2. Dec 7, 2011

### EternityMech

function goes to 0 when x>1

thats the answer.

3. Dec 7, 2011

### dacruick

Sorry, my first post was wrong, this is the edit. I think that if you do that, then x = 1, and is no longer satisfying the bounds of being between 1 and infinity.

4. Dec 7, 2011

### hamsterman

@EternityMech, would you elaborate? Is $1+\frac{1}{n} = 1$? I'm sure it isn't because raising both sides to the power of n gives $e = 1$

5. Dec 7, 2011

### dacruick

Yes I realized my error and changed my post.

EDIT: ooh and it looks like you realized that i realized my error and you also changed your post. well...thats good that this is all sorted out.

6. Dec 8, 2011

### vrmuth

its a lim sup of sequence of numbers or sequence of sets ? if numbers show me what are a1,a2,a3,.....and if sets show what are X1,X2,X3,......

7. Dec 8, 2011

### Sina

First of all, x is fixed. You can not replace it by a terms that depends on n, if you do, you change the sequence. What the sequence is completely determined by how n appears in it.

This is limsup so first you have to find supremum over m>n and then take the limit. I am assuming you are defining the supremum as follows (it may change but the general convention is this)

$\sup_{m>n} \left\{\frac{m}{x^m}:x\in\left( 1; \infty\right)\right\}$

However in our case the terms approach to zero (divide n+1th element by the nth element and take the limit, you will see that it is less than 1 so using theorems from calculus you can say that the limit exists and is zero). So when your sequence has a limit, liminf and limsup are equal to it and to each other (for instance on the contrary consider a sequence of points which converge to points on a sine graph, they do not have a limit but their limsup and liminf is 1 and -1). Thus the answer is zero.

More explicitly you can show that after a certain value of n (when we pass over the maximum of $\frac{m}{x^m}$) then
$sup_{m>n} \frac{m}{x^m} = \frac{n+1}{x^{n+1}}$ and so taking the limit gives you the result.

Last edited: Dec 8, 2011
8. Dec 8, 2011

### hamsterman

I guess I should add some context. I'm given $f_n(x)=\frac{n}{x^n}$. I need to analyze the convergence of this sequence of functions for all $x > 1$. First I see that $f(x) = \lim\limits_{n\rightarrow\infty} \frac{n}{x^n} = 0$. I then need to check whether $\forall\epsilon>0, \exists N,\forall n > N, \forall x: |f-f_n|(x) < \epsilon$ to know if the convergence is uniform. I do that using the expression in my first post. I'm sure that $x$ can be a function of $n$ because $\forall x$ goes after $\exists N$.

@vrmuth, I'm not sure how to define it. I guess it's a set of functions of n. Either way, the set is not discrete.

9. Dec 8, 2011

### Sina

well you should definitely distinguish between sequences of functions and sequences of points :) your first post looks more like a limit of sequence of points rather than functions. Still it is a bit strange to define x interms of n since it is just a free parameter no?

The limit fn(1+n) as n goes to infinity is not a point-wise limit but you just select the value of each fn at the point 1+n and look whether if that has a limit. I can't see what that would have to do with this question since we want to look at pointwise limits or uniform limit.

Last edited: Dec 8, 2011
10. Dec 8, 2011

### hamsterman

What I can't distinguish between is functions, numbers and sets. A function because it has an argument n, a number if n is a constant and a set if n is any natural number.

If limit fn(1+1/n) is not a pointwise limit, I assume that this sequence does not converge uniformly. In that case, my question is, for what x does it converge uniformly?

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