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Flipping signs for limits at negative infinity

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    [itex]\lim_{x \to -\infty} x + \sqrt{x^2 + 6x}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    Previous attempt was guessing it was [itex]\infty[/itex], but I see now my flaw and the actual answer is -3. Somewhere else on the web, might have been this forum, it was said that one could flip the sign and get

    [itex]\lim_{x \to -\infty} x + \sqrt{x^2 + 6x} = \lim_{x \to \infty} -x + \sqrt{x^2 + 6x}[/itex]

    Which I can see intuitively, since the what is under the radical would be positive either way, which implies that the sign only need be flipped for x. However, is there a generalized proof that includes any number of polynomials and roots, for this fact?

    - Farley
  2. jcsd
  3. Feb 8, 2009 #2


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    What you are doing is substituting -u for x. Then as x->-infinity, u->+infinity. But x+sqrt(x^2+6x) turns into -u+sqrt(u^2-6u), doesn't it?
  4. Feb 8, 2009 #3
    Yeah, that's probably a mistake.. But that does explain it. Thanks.
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