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Limit to negative infinity

  1. Jan 24, 2007 #1


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    1. The problem statement, all variables and given/known data

    limit as x-> -oo of x+(x^2+3)^(1/2) (square root)

    2. Relevant equations


    3. The attempt at a solution

    i first multiplied the top and bottom (which is just 1) by the conjugate to get:

    [x - (x^2+3)^(1/2)]

    then i divided by x on top and bottom to get:

    [x/x - (1 + 3/x2)^(1/2)]

    but now what do i do? the top goes to 0 but the bottom also goes to 0 (because its 1 - (1+0)^(1/2)). but the answer is suppose to be 0.

    0/0 isnt 0. can someone point me in the right direction? thanks.
  2. jcsd
  3. Jan 24, 2007 #2


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    Science Advisor
    Homework Helper

    [tex] \lim_{x\rightarrow -\infty} x+\sqrt{x^{2}+3} =\lim_{x\rightarrow \infty}
    \sqrt{x^{2}+3}-x [/tex]

    After multiplying by the conjugate expression you're facing the limit

    [tex] \lim_{x\rightarrow \infty} \frac{3}{\sqrt{x^{2}+3}+x} [/tex]

    which is trivial.

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