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Math Help - limits involving x-root functions

  1. #1
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    limits involving x-root functions

    Hey... This question is beyond me...

    find lim as x -> 0

    f(x) = 1 / ((x+1)^(1/x)) - 1/x

    any help would be greatly appreciated.
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  2. #2
    Super Member Bacterius's Avatar
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    f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}

    You want :

    \lim_{x\to 0} f(x)

    Therefore, you want :

    \lim_{x\to 0} f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}

    Let's not forget that \lim_{x\to 0}\frac{ 1}{x} = \pm \infty (depending on which side you come from)

    Consider if you are approaching 0 on the positive side :

    \lim_{x\to 0} f(x) = \frac{ 1}{1^{(+\infty)}} - (+\infty)

    Ugly, eh ?

    \lim_{x\to 0} f(x) = 1 - \infty

    Which by absorption (I'm not sure about the last part) :

    \lim_{x\to 0} f(x) = - \infty

    The other possibility (approaching 0 on the negative side) seems quite harder (since you get some weird results) but since I'm not really sure about what I've just done, I won't go on.
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  3. #3
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    Quote Originally Posted by Bacterius View Post
    Let's not forget that \lim_{x\to 0}\frac{ 1}{x} = \pm \infty (depending on which side you come from)
    This would mean \lim_{x \to 0}\frac{1}{x} does not exist.
    Last edited by Prove It; November 9th 2009 at 09:28 PM.
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  4. #4
    Super Member Bacterius's Avatar
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    What do you mean ? My maths teacher told us the inverse of x could have two different limits in 0 depending if you approach 0 from the negative side (the limit is -infinity) or the positive side (the limit is +infinity).
    Was he wrong or was I just doing something else when he said this ?

    EDIT : now that I think about it I remember he said something about 0, maybe he said \lim_{x \to 0}\frac{1}{x} = 0 ?
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  5. #5
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    Quote Originally Posted by Bacterius View Post
    What do you mean ? My maths teacher told us the inverse of x could have two different limits in 0 depending if you approach 0 from the negative side (the limit is -infinity) or the positive side (the limit is +infinity).
    Was he wrong or was I just doing something else when he said this ?

    EDIT : now that I think about it I remember he said something about 0, maybe he said \lim_{x \to 0}\frac{1}{x} = 0 ?
    1. Please note my edit, I wrote the wrong value for x to go to.

    2. He is not wrong, but if the function approaches the same value from the right as it does from the left, it does not exist.

    However, the left hand limit is -\infty and the right hang limit is \infty.
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  6. #6
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    Quote Originally Posted by Bacterius View Post
    f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}

    You want :

    \lim_{x\to 0} f(x)

    Therefore, you want :

    \lim_{x\to 0} f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}

    Let's not forget that \lim_{x\to 0}\frac{ 1}{x} = \pm \infty (depending on which side you come from)

    Consider if you are approaching 0 on the positive side :

    \lim_{x\to 0} f(x) = \frac{ 1}{1^{(+\infty)}} - (+\infty)

    Ugly, eh ?

    \lim_{x\to 0} f(x) = 1 - \infty

    Which by absorption (I'm not sure about the last part) :

    \lim_{x\to 0} f(x) = - \infty

    The other possibility (approaching 0 on the negative side) seems quite harder (since you get some weird results) but since I'm not really sure about what I've just done, I won't go on.

    Wow... looking at it that way makes it simple... thanks man.

    Also for the negative side its still easy...

    1/1^anything = 1 therefore answer is 1 - 1/x = 1 +infinity = infinity

    (or so I think)
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  7. #7
    Super Member Bacterius's Avatar
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    Yes you are right, it is basically the same thing on the other side, my mistake.
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