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Math Help - What is the limit of 1/(2^(n))

  1. #1
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    What is the limit of 1/(2^(n))

    As n approaches infinity.
    I tried using Squeeze Theorem but I got stuck...
    I said -1 < 1 < 1
    Then, -1/((2^n)) < 1/((2^n) < 1/((2^n) and I'm stuck...
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  2. #2
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    Quote Originally Posted by Mush89 View Post
    As n approaches infinity.
    I tried using Squeeze Theorem but I got stuck...
    I said -1 < 1 < 1
    Then, -1/((2^n)) < 1/((2^n) < 1/((2^n) and I'm stuck...
    It is completely transparent that 2^n gets very very large as n\to\infty.
    What does that tell us about \dfrac{1}{2^n}~?
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    1/(2^n) = 1/(infinity) = 0, but is it that simple? It's for a calc. class and this is the limit of a sequence...
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    Quote Originally Posted by Mush89 View Post
    1/(2^n) = 1/(infinity) = 0, but is it that simple? It's for a calc. class and this is the limit of a sequence...
    NEVER, NEVER write that. It is an automatic F.
    Now are you asked to prove the limit is 0?
    Or are you simply asked to find the limit?
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    I was taught that 1 is a constant, and that c/(infinity) = 0.
    And it says, "Does the limit exist, and if so, what is it's value? Give detailed explanation."
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    Quote Originally Posted by Mush89 View Post
    I was taught that 1 is a constant, and that c/(infinity) = 0.
    And it says, "Does the limit exist, and if so, what is it's value? Give detailed explanation."
    I truly that you were not actually taught that.
    But then again there are a lot of calculus teachers that have no business being in any mathematics class room.
    Last edited by Plato; April 5th 2011 at 03:48 PM.
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  7. #7
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    Quote Originally Posted by Mush89 View Post
    I was taught that 1 is a constant, and that c/(infinity) = 0.
    And it says, "Does the limit exist, and if so, what is it's value? Give detailed explanation."
    I expect you were told that \lim_{x \to \infty}\frac{c}{x} = 0. You can NOT ever substitute \displaystyle \infty as though it's a number. It's more of an idea.
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