Limit of Function x→∞: 0<a<1, a>1

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Homework Help Overview

The discussion revolves around finding the limit of the function as x approaches infinity for two cases: when 0 < a < 1 and when a > 1. The mathematical expression being analyzed is 〔 (a^x - 1 )/(a - 1) 〕^1/x.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the limit by rewriting the expression and considering the behavior of the function as x approaches infinity. Some participants express uncertainty about their attempts, while others suggest specific cases to analyze first.

Discussion Status

There is ongoing exploration of the limit for the case when 0 < a < 1, with some participants questioning the correctness of their reasoning. The discussion has not reached a consensus, particularly regarding the case when a > 1, which remains open for further exploration.

Contextual Notes

Participants note the importance of showing their efforts in the problem-solving process, and there is a recognition of the need to clarify assumptions about the values of a.

chiakimaron
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I have no idea of the following question, please help me

Find limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

for (1) 0<a<1
(2) a>1
 
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what have you tried? If you are not showing your effort, then you can't get any help
 
let y=〔 (a^x - 1 )/(a - 1) 〕^1/x

Iny=1/x In(a^x - 1 )/(a - 1)

then I can't do it
 
start with 0<a<1 then what happens?
 
Last edited:
i really don't know
 
limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

= [(-1)/(a-1)]^0 = 1

is it correct?
 
chiakimaron said:
let y=〔 (a^x - 1 )/(a - 1) 〕^1/x

Iny=1/x In(a^x - 1 )/(a - 1)

then I can't do it

Keep going! Let x\to\infty

\lim_{x\to\infty}\frac{\ln\left(\frac{a^x-1}{a-1}\right)}{x}=\frac{\lim_{x\to\infty}\ln\left(\frac{a^x-1}{a-1}\right)}{\lim_{x\to\infty}x}

and remember ln(-1/(a-1)) is some constant.
 
chiakimaron said:
limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

= [(-1)/(a-1)]^0 = 1

is it correct?

yes! :-)

And what if a>1 ?
 

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