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

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To prove that lim [(an+b)^(1/n) - 1] = 0 as n approaches infinity for positive real numbers a and b, the discussion suggests using L'Hopital's rule after rearranging the expression. Taking the natural logarithm leads to an indeterminate form, allowing the application of L'Hopital's rule to evaluate the limit. However, it is clarified that L'Hopital's rule is applicable to functions rather than sequences, prompting a rephrasing of the question to focus on sequences. The conversation emphasizes that if a sequence can be expressed as a function, L'Hopital's rule can be utilized effectively. Ultimately, the participants agree on the validity of treating certain sequences as functions for limit evaluation.
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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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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