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Math Help - limit proof

  1. #1
    Senior Member BAdhi's Avatar
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    limit proof

    I want to check whether I've proved the following limit question correctly. Because i don't feel so.
    <br />
\lim_{x \to \infty} 3^{\frac{1}{x}}

    attempt:

    let \epsilon >0 and a>0

    x > a \implies \frac{1}{x}<\frac{1}{a} \implies 3^{\frac{1}{x}}-1<3^{\frac{1}{a}}-1

    so define a=\frac{1}{\log_3(\epsilon+1)}

    then x>a \implies 3^{\frac{1}{x}}-1<\epsilon \implies |3^{\frac{1}{x}}-1|<\epsilon

    so can anyone check my work on this
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  2. #2
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    It looks fine.
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  3. #3
    Senior Member BAdhi's Avatar
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    yet i can prove \lim_{x \to \infty} 3^{\frac{1}{x}} = 0 in the same way?
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  4. #4
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    Quote Originally Posted by BAdhi View Post
    yet i can prove \lim_{x \to \infty} 3^{\frac{1}{x}} = 0 in the same way?
    I doubt it...

    You can prove it in other way if you don't want to use the formal definition...

    \displaystyle \lim_{x \to \infty}3^{\frac{1}{x}} = \lim_{x \to \infty}e^{\ln{\left(3^{\frac{1}{x}}\right)}}

    \displaystyle = \lim_{x \to \infty}e^{\frac{1}{x}\ln{3}}

    \displaystyle = e^{\lim_{x \to \infty}\frac{\ln{3}}{x}}

    \displaystyle = e^0

    \displaystyle = 1.
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  5. #5
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    log_3(\epsilon +1)>0 is important. For small \epsilon, log_3(\epsilon )<0 which is where the argument fails for 0.
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  6. #6
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    Also, in your argument you need to look at 3^\frac{1}{x}-1>-\epsilon more carefully. It works out fine here, but this is where the argument might fall apart if you try to prove the limit is equal to something larger than 1.

    The important thing here is that 1-\epsilon is always less than 1, so that you always get a negative number when you take the log (and 1 is the largest number that has this property for all \epsilon).
    Last edited by DrSteve; December 28th 2010 at 06:34 AM.
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