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Math Help - Limit w. conjugate/ L'Hopital's Rule

  1. #1
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    Post Limit w. conjugate/ L'Hopital's Rule

    our instructor gave us this problem as a heads up because it will be on our quiz tomorrow, but it is pretty tough. He told us to multiply by the conjugate...

    lim (x^2+x)^1/2 + x
    x -> (infinite)

    I have determined that the conjugate is (x^2+x)^1/2 - x, but when i do that, and then to L'hopital's, i feel as tho continuing will get me nowhere... help?
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    Hello,
    Quote Originally Posted by Smac View Post
    our instructor gave us this problem as a heads up because it will be on our quiz tomorrow, but it is pretty tough. He told us to multiply by the conjugate...

    lim (x^2+x)^1/2 + x
    x -> (infinite)

    I have determined that the conjugate is (x^2+x)^1/2 - x, but when i do that, and then to L'hopital's, i feel as tho continuing will get me nowhere... help?
    for the conjugate !

    Now, multiply and divide by the conjugate :

    \sqrt{x^2+x}+x=(\sqrt{x^2+x}+x) \times \frac{\sqrt{x^2+x}-x}{\sqrt{x^2+x}-x}

    =\frac{(\sqrt{x^2+x})^2-x^2}{\sqrt{x^2+x}-x}

    =\frac{x}{\sqrt{x^2+x}-x}

    You can apply l'Hospital's, but there are nicer methods :

    =\frac{x}{\sqrt{x^2\left(1+\frac 1x\right)}-x}=\frac{x}{x \sqrt{1+\frac 1x}-x}

    =\frac{x}{x \left(\sqrt{1+\frac 1x}-1\right)}=\frac{1}{\sqrt{1+\frac 1x}-1}}

    LImit to infinity : \sqrt{1+\frac 1x} \to 1, with superior values to 1. So \sqrt{1+\frac 1x}-1 \to 1 with positive values.

    Thus the limit is...


    (you would have found the same with l'Hospital's)

    --------------------------------------------------------
    My point is that the limit would have been more friendly if it was (x+x)^(1/2) - x.
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  3. #3
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    wow....that was easy as hell <----stupid
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