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Math Help - Find Limit

  1. #1
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    Find Limit

    Please show work
    Find Limit-limit.png
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  2. #2
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    Re: Find Limit

    factorize the denominator as (sqrx-1)(sqrx+1) then simplify and the limit is in front of you.........1/2
    MINOAS
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  3. #3
    Senior Member Educated's Avatar
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    Re: Find Limit

    You can use l'hopital's rule, since you get 0/0 by direct substitution.

    Which says:
    \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}
    as long as the first limit is 0/0 or infinity/infinity


    Therefore you get:
    \lim_{ x \to \1} \frac{ \sqrt{x} - 1}{x - 1} = \lim_{x\to\1} \frac{\frac{1}{2 \sqrt{x}}}{1} = \dfrac{1}{2}
    Last edited by Educated; April 14th 2013 at 09:58 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Find Limit

    Quote Originally Posted by leonhart View Post
    Please show work
    Click image for larger version. 

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    As MINOANMAN said, but let's do a quick substitution to make it look a bit nicer. Let x = y^2. Then we have
    \lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \lim_{y \to 1} \frac{y - 1}{y^2 - 1} = \lim_{y \to 1} \frac{y - 1}{(y - 1)(y + 1)}

    etc.

    -Dan
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  5. #5
    Junior Member Bradyns's Avatar
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    Re: Find Limit

    That is a rather elegant solution.
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