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Thread: yet another limit proof

  1. #1
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    yet another limit proof

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

    Are you familiar with this.
    Last edited by mr fantastic; Oct 19th 2009 at 09:11 PM. Reason: Corrected a typographical omission.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Convergence and the $\displaystyle \epsilon-N$ criterion: $\displaystyle \lim_{n \rightarrow +\infty} S_n= L$ iff for all $\displaystyle \epsilon > 0$ there exists $\displaystyle N > 0$ such that n>N implies $\displaystyle |S_n - L| < \epsilon$.

    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 $\displaystyle 0<a<1$, let $\displaystyle N=\log_a\epsilon$. Therefore,

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

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

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