
#1
Jun2603, 05:17 PM

P: 321

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 ab<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. 



#2
Jun2603, 05:43 PM

Emeritus
Sci Advisor
PF Gold
P: 5,540

First, rearrange it to: lim(an+b)^{1/n}=1 n>∞ Then take the natural log of both sides to get: lim ln(an+b)/n=0 n>∞ This goes to ∞/∞, which is an indeterminate form and ripe for L'Hopital's rule. 



#3
Jun2603, 05:58 PM

P: 321

LOL, thanks Tom and L'hopital
lim ln(an+b)/n n>[oo] = lim a/(an+b) n>[oo] =0 



#4
Jun2703, 04:42 AM

P: 321

Limit of a sequence
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 {a_{n}} 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) 



#5
Jun2703, 06:56 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

It is true that L'hopital's rule applies to functions rather than sequences.
However, IF we can convert a sequence a_{n} 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, a_{n}> L. The other way doesn't necessarily work the function might not have a limit, depending on how it is defined for noninteger values. 



#6
Jun2703, 07:13 AM

P: 321




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