show the limit is b

**wopashui** · March 21st 2010, 09:53 AM

show that if $0<a<b, then (a^n + b^n)^{1/n} --> b$

**Plato** · March 21st 2010, 09:59 AM

Originally Posted by wopashui

show that if $0<a<b, then (a^n + b^n)^{1/n} --> b$

$b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{b^n + a^n }} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2}$

**wopashui** · March 21st 2010, 10:24 AM

Originally Posted by Plato

$b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{b^n + a^n }} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2}$

this just proved it's bounded, is it?

**drumist** · March 21st 2010, 11:03 AM

Originally Posted by wopashui

this just proved it's bounded, is it?

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

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

What does that tell you then if $b \le \sqrt[n]{a^n + b^n} \le b\sqrt[n]{2}$ ?

**chiph588@** · March 21st 2010, 11:40 AM

Originally Posted by wopashui

this just proved it's bounded, is it?

Apply the squeeze theorem.

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