Simple analysis proof: x^n -> a^n as x -> a

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Homework Help Overview

The problem involves proving that \( x^n \) approaches \( a^n \) as \( x \) approaches \( a \), focusing on the behavior of the function as the variable approaches a specific value.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to establish a relationship between \( |x-a| \) and \( |x^n-a^n| \) through the epsilon-delta definition of limits. There is mention of factoring the expression \( x^n - a^n \) and exploring the implications of the factorization. Questions arise regarding the assumptions about the variable \( n \) and its constraints.

Discussion Status

The discussion is active, with participants providing insights into the factorization approach and questioning the assumptions regarding \( n \). Some guidance has been offered on how to relate the expressions, but there is no explicit consensus on the method to show the desired inequality.

Contextual Notes

There is a noted ambiguity regarding whether \( n \) is a positive integer, which affects the assumptions made in the problem. Participants are also considering the implications of this assumption on the proof.

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Homework Statement


Prove that x^n approaches a^n as x approaches a.


Homework Equations





The Attempt at a Solution


I understand the concept here ... need to find a delta>0 for epsilon>0 s.t. |x-a|<delta implies |x^n-a^n|<epsilon. For some reason I can't solve this one. Thanks.
 
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You know you can factor x^n-a^n=(x-a)*(x^(n-1)+x^(n-2)*a+ ... + a^(n-1)), right?
 
Oh! Somehow I missed that. Thank you very much for your help.
 
Dick assumes n is a positive integer, which you did not tell us under "all variables and given/known data" ... so are we to assume n is NOT know to be a positive integer?
 
still, how do you show that (x^(n-1)+x^(n-2)*a+ ... + a^(n-1)) is < epsilon?
 
You don't, and no one has claimed it is. You want [itex]|x^n- a^n|= |x- a||x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|< \epsilon[/itex] for |x- a| small enough. That is the same as choosing [itex]|x- a|< \epsilon/|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|[/itex] and you can do that by finding any positive lower bound on [itex]|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|[/itex] for x close to a.
 

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