Understanding the Limit of xn as n Tends to Infinity

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    Infinity Limit
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Homework Help Overview

The discussion revolves around finding the limit of the expression \( x_n = (n^2 + \exp(n))^{1/n} \) as \( n \) approaches infinity. The subject area includes limits and exponential functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various methods to determine the limit, including the use of logarithms and l'Hôpital's rule. There is uncertainty about the applicability of certain techniques and the dominance of terms within the expression.

Discussion Status

Participants are actively discussing different approaches, with some suggesting the use of logarithmic properties and others expressing concerns about the limitations of their methods. There is a recognition of the need to analyze the behavior of the terms as \( n \) increases, particularly the relationship between \( n^2 \) and \( \exp(n) \).

Contextual Notes

Some participants question whether the use of l'Hôpital's rule is permissible, indicating potential constraints on the methods allowed for the homework assignment.

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Homework Statement



Find the limit as n tends to infinity of xn = (n^2 + exp(n))^(1/n)

Homework Equations



maybe use ( 1 + c/n )^n tends to exp(c)

The Attempt at a Solution



I know that inside the barckets are both inceasing and the 1/n makes it decrease but how do i find out which is stronger and what the limit is?
 
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Try finding the limit of the ln of the function.

Tip: ln(a^b) = b ln(a)
 
i already tried that and i can't see how it helps, (1/n)ln(n^2 + exp(n)) has the same problem...
 
Try l'Hopital's rule, if you know that.
 
i have a feeling I am not allowed to use it, is there another way?
 
What you really need to know is that n^2/exp(n)->0 as n->infinity. There are a variety of ways to show that - try to think of one. Once you done that then ln(exp(n)+n^2)=ln(exp(n)*(1+n^2/exp(n))=ln(exp(n))+ln(1+n^2/exp(n)) etc.
 
[tex](n^2+e^n)^{1/n} = e(1+n^2/e^n)^{1/n}[/tex] now it is trivial but the fact that [tex](1+x_n/n)^{1/n} \to e^x[/tex] if [tex]x_n\to x[/tex].
 

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