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Math Help - show the limit is b

  1. #1
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    show the limit is b

    show that if 0<a<b, then (a^n + b^n)^{1/n} --> b
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  2. #2
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    Quote Originally Posted by wopashui View Post
    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}
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  3. #3
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    Quote Originally Posted by Plato View Post
    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?
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  4. #4
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    Quote Originally Posted by wopashui View Post
    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} ?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by wopashui View Post
    this just proved it's bounded, is it?
    Apply the squeeze theorem.
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