Trivial limit ( 1 - (-x)^n ) / 1 + x

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (1 - (-x)^n) / (1 + x) as n approaches infinity. The subject area involves limits and properties of exponential functions, particularly in the context of variable x and its constraints.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster expresses confusion regarding the application of limit rules and questions the reasoning behind the limit equating to 1 / (1 + x). Some participants suggest that conditions on x are necessary for the limit to hold true, specifically within the range of -1 < x < 1. Others discuss the behavior of x raised to the power of n as n approaches infinity.

Discussion Status

Participants are exploring the conditions under which the limit can be evaluated. Some guidance has been provided regarding the necessary constraints on x, and there is an acknowledgment of understanding among participants about the implications of these constraints.

Contextual Notes

There is an emphasis on the range of x being critical for the evaluation of the limit, specifically noting that x must be between -1 and 1 for the limit to behave as discussed.

davitz
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Homework Statement



lim ( 1 - ( - x ) ^ n ) / ( 1 + x ) as n -> infinity

Homework Equations



I can't understand why this equals to 1 / ( 1 + x ) (No matter what power of " n " was x e.q: x ^ 2n or x ^ ( n ^ 2 )

The Attempt at a Solution



I have no clue what rule to apply. I thought it might be a case of using the lim ( 1 + 1 / n ) ^ n to get to " e " but this seems like a non-sense in this case.
 

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I think you first need to have a condition on x before that is true.

0 < x < 1, right?

If so, anything between 0 and 1 raised to a power of infinity will tend to 0, as it gets smaller with each successive multiplication.

[tex]\lim_{n\to \infty} x^n = 0[/tex]

where -1<x<1
 
Thank so much for the reply! Yes, x > -1 and x < 1 or -1 < x < 1 and now I understand why this is the result!

Thank you again!
 
Glad to have been of help! :smile:
 

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