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Thread: limits

  1. #1
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    limits- please help its very urgent...

    please see the attatchment
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  2. #2
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    As x gets indefinetly large 1/x will get closer and closer to zero and any non zero real number rasied to the zero power yeilds one so the lim as x-->infinity of (x)^(1/x) = 1
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    please see the attatchment
    You are asking - what is:

    $\displaystyle Lt_{x\rightarrow \infty} \sqrt[x] x$

    with a choice of answers $\displaystyle \infty,\ 1,\ 0.$

    Consider:

    $\displaystyle log(Lt_{x\rightarrow \infty} \sqrt[x] x)$,

    which is equal to:

    $\displaystyle Lt_{x\rightarrow \infty}log( \sqrt[x] x)\ =\ Lt_{x\rightarrow \infty} \frac{log(x)}{x}$.

    Now if you know that $\displaystyle log(x)$ grows more slowly than any
    positive power of $\displaystyle x$ you are done, as:

    $\displaystyle Lt_{x\rightarrow \infty} \frac{log(x)}{x}\ =\ 0$,

    and so:

    $\displaystyle Lt_{x\rightarrow \infty} \sqrt[x] x\ =\ 1$.

    If you don't know that $\displaystyle log(x)$ grows more slowly than any
    positive power of $\displaystyle x$, then you can get the same result by a slightly
    convoluted application of L'Hopital rule.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by mathproblem
    As x gets indefinetly large 1/x will get closer and closer to zero and any non zero real number rasied to the zero power yeilds one so the lim as x-->infinity of (x)^(1/x) = 1
    Would not the same argument apply to $\displaystyle \sqrt[x]{x^x}$?

    RonL
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