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Math Help - Evaluating limits...

  1. #1
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    Evaluating limits...

    \lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}

    I can't seem to evaluate it properly...

    I don't know how to use the math tag tool so I had to copy and tinker with it until I had what I needed... The edited equation is what I'm having difficulty with.
    Last edited by Neversh; January 31st 2010 at 11:26 PM. Reason: The equation was difficult to understand I guess. :(
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Neversh View Post
    Lim ( 1/(x([1 + x]^1/2) -1/x)
    x->0

    I can't seem to evaluate it properly...
    Let's assume the limit exists and equals L\ne 0. Then, \frac{1}{L}=\lim_{x\to0}\frac{1}{\frac{1}{x\sqrt{1  +x}}-\frac{1}{x}}=\lim_{x\to0}\frac{x\sqrt{1+x}\cdot x}{x-x\sqrt{1+x}}=\lim_{x\to0}\frac{x\sqrt{1+x}}{1-\sqrt{1+x}} =\lim_{x\to0}\frac{x}{\frac{1}{\sqrt{1+x}}-1}=\frac{1}{f'(0)} where f(x)=\frac{1}{\sqrt{1+x}}
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by Neversh View Post
    lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}

    The answer is - 1/2. I can't seem to get this answer
    Start by making a common denominator.
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  4. #4
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    Quote Originally Posted by Neversh View Post
    \lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}

    The answer is - 1/2. I can't seem to get this answer
    \lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}=\lim_{x\to0}{\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}}
    =\lim_{x\to0}{\frac{(1-\sqrt{1+x})(1+\sqrt{1+x})}{(x\sqrt{1+x})(1+\sqrt{1  +x})}}
    =\lim_{x\to0}{\frac{(-x)}{(x\sqrt{1+x})(1+\sqrt{1+x})}}
    =\lim_{x\to0}{\frac{(-1)}{(\sqrt{1+x})(1+\sqrt{1+x})}}
    =\frac{-1}{2}
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