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Math Help - limits question

  1. #1
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    limits question

    explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.



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  2. #2
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    Quote Originally Posted by alpha View Post
    explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.

    i can see it on the graph that i have sketched but i can't explain it.

    Thanks.
    The graph itself is the best answer for your question.
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  3. #3
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    Quote Originally Posted by alpha View Post
    explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.



    Thanks.
    hi
    Hint,
    2^{\frac{1}{x}}=\exp(\log(2^{\frac{1}{x}}))
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  4. #4
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    Quote Originally Posted by Raoh View Post
    hi
    Hint,
    2^{\frac{1}{x}}=\exp(\log(2^{\frac{1}{x}}))
    i can see it's the same graph but what is the explanation for this.

    thanks.
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  5. #5
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    Quote Originally Posted by alpha View Post
    i can see it's the same graph but what is the explanation for this.

    thanks.
    \lim_{x\to 0^{+}}\log(2^{\frac{1}{x}})=\lim_{x\to 0^{+}}\frac{1}{x}\log(2)=\infty
    \lim_{x\to 0^{-}}\log(2^{\frac{1}{x}})=\lim_{x\to 0^{-}}\frac{1}{x}\log(2)=-\infty
    Remember,
    e^{\infty }=\infty and e^{-\infty }=0
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  6. #6
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    x=1,\ 2^{\frac{1}{x}}=2^1=2

    x=0.5,\ 2^{\frac{2(0.5)}{0.5}}=2^2

    x=0.1,\ 2^{\frac{10(0.1)}{0.1}}=2^{10}

    As\ x\ goes\ to\ zero\ from\ above,\ f(x)\ goes\ to\ infinity.

    x=-1,\ f(-1)=2^{-1}=\frac{1}{2}

    x=-0.5,\ f(-0.5)=2^{-2}=\frac{1}{2^2}

    x=-0.1,\ f(-0.1)=2^{-10}=\frac{1}{2^{10}}

    As\ x\ goes\ to\ zero\ from\ below,\ f(x)\ goes\ to\ zero.
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