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Asymptote Problem

  1. Jun 5, 2004 #1
    My problem asks to find the vertical asymptotes of x^3/(x^2+3x-10)

    I found –5 & 2 to be vertical asymptotes but what I can’t figure out is
    how as x->-5- = -oo and x->-5+ = +oo

    I have calculated lim x->-5+ x^3/(x^2+3x-10) = -125/(0)(-5 - -2) = -125/-0 = +oo

    But I don’t see –oo lim x->+5+ x^3/(x^2+3x-10) = ???????????????

    Can you show me the algebra and how the signs work out to give me –oo

  2. jcsd
  3. Jun 5, 2004 #2

    I think i have it ...

    lim x-> -5+ x^3/(x^2+3x-10) = +oo since x^3/(x+5)(x-2) > 0 for x > -5

    and lim x->-5- x^3/(x^2+3x-10) = -oo since x^3/(x+5)(x-2) < 0 for x < -5
  4. Jun 5, 2004 #3


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    Didn't look at the actual math, but that's the general idea. You can nomrally if it's going to approach an infinity, then look at the signs to tell whether it would be positive or negative.
  5. Jun 5, 2004 #4


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    Homework Helper

    Looking at the graph is always (almost) helpful.
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