Re: Limit

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 03/24/05

Date: Thu, 24 Mar 2005 05:40:47 -0600

On 24 Mar 2005 02:45:36 -0800, sweet_sorrow30@yahoo.com (TT) wrote:

>a,b > 0 .
>
>Please evaluate lim{ [(a^(1/k) + b^(1/k)]/2 }^k as k goes to infinite.

That would be sqrt(ab).

Note that

e^t = 1 + t + E(t),

where |E(t)/t^2| is bounded for |t| <= 1.
Also note that

log(1+t) = t + L(t),

where |L(t)/t^2| is bounded for |t| <= 1/2.

Assume that ab = 1 (when you hand this in make certain
to explain why it's legal to assume this.) Then
log(a) + log(b) = 0, hence

(a^(1/k) + b^(1/k)]/2 = (xp(log(a)/k) + exp(log(b)/k))/2

= 1 + H(k),

where |H(k)|<= c/k^2. So

log([(a^(1/k) + b^(1/k)]/2 }^k)

= k log(H(k)),

hence

|log([(a^(1/k) + b^(1/k)]/2 }^k)| <= ck/k^2

for large k.

>Thanks!

************************

David C. Ullrich