show the limit is b

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March 21st 2010, 09:53 AM
wopashui

show the limit is b

show that if $0<a<b, then (a^n + b^n)^{1/n} --> b$
March 21st 2010, 09:59 AM
Plato

Quote:

Originally Posted by wopashui

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

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

Quote:

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?
March 21st 2010, 11:03 AM
drumist

Quote:

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 $b \le \sqrt[n]{a^n + b^n} \le b\sqrt[n]{2}$ ?
March 21st 2010, 11:40 AM
chiph588@

Quote:

Originally Posted by wopashui

this just proved it's bounded, is it?

Apply the squeeze theorem.

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