Question

Compute the limit of this sequence:

x=(a^n + b^n)^(1/n)

where a and b are real postive numbers!

Answer 1

Factor out a^ to get (a^n)^1/n (1+(b/a)^n)^1/n.

The first factor is just a, and then you can use L'hopitals rule to evaluate the second.

Answer 2

Don't you have to evaluate when b<a, b>a and a=b??

Answer 3

I don't think that matters...

Answer 4

\[\exp\left(\lim_{n\to\infty}{\ln(a^n+b^n)\over n}\right)=\exp\left(\lim_{n\to\infty}{na^{n-1}+nb^{n-1}\over a^n+b^n}\right)=\]maybe you are right...

Answer 5

You have to factor out the larger of a and b so that the second term vanishes as n gets big.

\[(a^n + b^n)^{1/n} = (a^n)^{1/n} (1+ (b/a)^n)^{1/n} = a

When a=b, you get a, since 2^{1/n} goes to 1.

Answer 6

Ok, let's see, you factor out a when a>b so you get the sequence --> a
now if a=b, (2a^n)^(1/n)=a(2)^(1/n) so the sequence --> a , too
we almost prove that the limit is a
but when b>a, if we factor out b,
b[(a/b)^n + 1]^(1/n) the sequence --> b
That's what I don't understand, on this case, the limit doesn't have to be a as well??

Answer 7

Given two numbers a and b, only one of them is going to be largest, so the limit is unique
in every case. - just not the same in different cases.

I guess you could call the limit = max(a,b) in order to cover all the cases.....

Answer 8

yeah you're right, thanx (: