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Math Help - yet another limit proof

  1. #1
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    yet another limit proof

    how would I show that \lim_{x\to \infty}\frac{2^x}{3^{2x+1}} exists and what it is?
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  2. #2
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    Quote Originally Posted by binkypoo View Post
    how would I show that \lim_{x\to \infty}\frac{2^x}{3^{2x+1}} exists and what it is?
    The expression you're taking the limit of can be written \frac{1}{3} \cdot \left(\frac{2}{3^2}\right)^x and the limiting value should be apparent. Now I suppose you might have to give an epsilon-delta proof of this limit.
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  3. #3
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    more generally, how could I prove that \lim_{x\to \infty} a^x = 0 when 0\leq a \leq 1?
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    Quote Originally Posted by dannyboycurtis View Post
    more generally, how could I prove that \lim_{x\to \infty} a^x = 0 when 0\leq a \leq 1?
    Convergence and the \epsilon-N criterion: \lim_{n \rightarrow +\infty} S_n= L iff for all \epsilon > 0 there exists N > 0 such that n > N implies |S_n - L| < \epsilon.

    Are you familiar with this.
    Last edited by mr fantastic; October 19th 2009 at 10:11 PM. Reason: Corrected a typographical omission.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Convergence and the \epsilon-N criterion: \lim_{n \rightarrow +\infty} S_n= L iff for all \epsilon > 0 there exists N > 0 such that n>N implies |S_n - L| < \epsilon.

    Are you familiar with this.
    I added a missing part to the definition above.
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  6. #6
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    the problem Im having is finding an appropriate N.
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    the problem Im having is finding an appropriate N.
    For 0<a<1, let N=\log_a\epsilon. Therefore,

    n>N \implies a^n<a^N=a^{\log_a\epsilon}=\epsilon
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  8. #8
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    Quote Originally Posted by redsoxfan325 View Post
    For 0<a<1, let N=\log_a\epsilon. Therefore,

    n>N \implies a^n<a^N=a^{\log_a\epsilon}=\epsilon
    Actually, one small adjustment: since \log_a\epsilon is unlikely to be a natural number, you need to take N=\lceil\log_a\epsilon\rceil, where \lceil x\rceil is the smallest integer greater than x. Then you have basically the same thing with one tiny adjustment (in red):

    n>N \implies a^n<a^N{\color{red}<}a^{\log_a\epsilon}=\epsilon
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