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Math Help - limit problem

  1. #1
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    Question limit problem

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    Last edited by Revelsyn; October 13th 2008 at 07:22 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Revelsyn View Post
    lim u --> 1 (sqrt1+u) - (sqrt1+u^2)/ (u-1)

    i tried to solve this many times without success, and i have a feeling it's very obvious ><

    (the answer is -(sqrt2)/4 )
    do you mean \frac {\sqrt{1 + u} - \sqrt{1 + u^2}}{u - 1} ?

    if so, multiply by \frac {\sqrt{1 + u} + \sqrt{1 + u^2}}{\sqrt{1 + u} + \sqrt{1 + u^2}} (the conjugate of the numerator over itself) and simplify as far as possible before taking the limit. you should be able to cancel out the denominator and hence remove the problem
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  3. #3
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    Rationalize the numerator:
    \lim_{u \to 1} \left( \frac{\sqrt{1+u} - \sqrt{1+u^2}}{u-1} \cdot \frac{{\color{red}\sqrt{1+u} + \sqrt{1+u^2}}}{{\color{red}\sqrt{1+u} + \sqrt{1+u^2}}}\right)

    Simplify the numerator (notice it is a difference of squares and don't worry about the denominator .. you'll see why)
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    Last edited by Revelsyn; October 13th 2008 at 07:22 AM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Revelsyn View Post
    then i did :
    =(1+ u) - (1 + u^2)/(u-1)[(sqrt1+u) + (sqrt1+u^2)]

    =u - u^2/ (u-1)[(sqrt1+u) + (sqrt1+u^2)]

    =-u(u-1)/(u-1)[(sqrt1+u) + (sqrt1+u^2)]

    =-1/2(sqrt2) ???
    yes

    you are correct

    (note that you have to rationalize the fraction to make the answer look like the one you want)
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    Last edited by Revelsyn; October 13th 2008 at 07:22 AM.
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