# show the limit is b

• Mar 21st 2010, 09:53 AM
wopashui
show the limit is b
show that if $\displaystyle 0<a<b, then (a^n + b^n)^{1/n} --> b$
• Mar 21st 2010, 09:59 AM
Plato
Quote:

Originally Posted by wopashui
show that if $\displaystyle 0<a<b, then (a^n + b^n)^{1/n} --> b$

$\displaystyle b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{b^n + a^n }} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2}$
• Mar 21st 2010, 10:24 AM
wopashui
Quote:

Originally Posted by Plato
$\displaystyle b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{b^n + a^n }} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2}$

this just proved it's bounded, is it?
• Mar 21st 2010, 11:03 AM
drumist
Quote:

Originally Posted by wopashui
this just proved it's bounded, is it?

What is $\displaystyle \lim_{n\to\infty} b$ ?

What is $\displaystyle \lim_{n\to\infty} b \sqrt[n]{2}$ ?

What does that tell you then if $\displaystyle b \le \sqrt[n]{a^n + b^n} \le b\sqrt[n]{2}$ ?
• Mar 21st 2010, 11:40 AM
chiph588@
Quote:

Originally Posted by wopashui
this just proved it's bounded, is it?

Apply the squeeze theorem.