Prove Limit of Sequence: a, b, (an+b)1/n-1 = 0

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Homework Help Overview

The discussion revolves around proving the limit of a sequence defined as \((an+b)^{1/n} - 1\) as \(n\) approaches infinity, where \(a\) and \(b\) are positive real numbers. Participants are exploring methods to demonstrate that this limit equals zero.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest using L'Hopital's rule after transforming the limit into a logarithmic form. Others question the applicability of L'Hopital's rule to sequences versus functions, leading to a discussion about the conditions under which a sequence can be treated as a function.

Discussion Status

The conversation is ongoing, with participants sharing different perspectives on the use of L'Hopital's rule and clarifying the nature of the sequence involved. There is no explicit consensus yet, but various approaches are being considered.

Contextual Notes

Participants note the distinction between sequences and functions, particularly in relation to the applicability of L'Hopital's rule. There is also an acknowledgment of the need to rephrase the original question to clarify that it pertains to a sequence.

KLscilevothma
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It isn't a homework problem but I think I better post it here instead of Mathematics forum, since it belongs to "exam help".

Prove that for any positive real numbers a and b,
lim [(an+b)1/n-1] = 0
n->inf

I don't need to use things like |a-b|<epsilon. A simple way will do. I know it's an easy question but I don't know where to start. Could someone please help.
 
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Originally posted by KL Kam
Prove that for any positive real numbers a and b,
lim [(an+b)1/n-1] = 0
n->inf

This one just screams "L'Hopital!"

First, rearrange it to:

lim(an+b)1/n=1
n-->&infin;

Then take the natural log of both sides to get:

lim ln(an+b)/n=0
n-->&infin;

This goes to &infin;/&infin;, which is an indeterminate form and ripe for L'Hopital's rule.
 
LOL, thanks Tom and L'hopital

lim ln(an+b)/n
n->[oo]

= lim a/(an+b)
n->[oo]
=0
 
Oh sorry, I forgot to mention
(an+b)1/n-1
is a sequence, not a function. I think L'hopital's rule applies to differentiable functions only.

Perhaps I better rephase the question a bit.
A sequence {an} is defined by (an+b)1/n-1
Prove that
lim (an+b)1/n-1 = 0
n->inf
(a and b are real numbers and n is a positive integer)
 
It is true that L'hopital's rule applies to functions rather than sequences.


However, IF we can convert a sequence an to a function f(x) (we can't if the sequence involves things like n! or "floor" or "ceiling" that can't be written simply as a continuous function), then f(x)-> L, an-> L. The other way doesn't necessarily work- the function might not have a limit, depending on how it is defined for non-integer values.
 
Originally posted by HallsofIvy
However, IF we can convert a sequence an to a function f(x) (we can't if the sequence involves things like n! or "floor" or "ceiling" that can't be written simply as a continuous function), then f(x)-> L, an-> L. The other way doesn't necessarily work- the function might not have a limit, depending on how it is defined for non-integer values.
So we can treat a sequence as a function if it is an "elementary" one like the one I posted, and can apply L'hopital's rule, is it correct?
 

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