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.
This one just screams "L'Hopital!" 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.
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)
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 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?