# Limit of a sequence

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.

Tom Mattson
Staff Emeritus
Gold Member
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)

HallsofIvy