1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Flipping signs for limits at negative infinity
Loading...