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Homework Help: More limits problems

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data
    1. limx->-inf (sqrt(x2+6x-1) + x)
    2. Let g(x) = x10/9, find limh->0 g((209+h) - g(209))/h

    2. Relevant equations
    None that I know of.

    3. The attempt at a solution
    1. Well, first I put the -infinty in and I think it's an indetermination because it's inf-inf. So I decided to rationalize the equation and got:
    sqrt(x2+6x-1) -x)
    I figured the limit of the denominator has to be +infinity, limit of the top is -infinity, which would result in getting a -infinity. But the answer says it's -3.

    2. I substituted and got this:
    (209+h)10/9 - 209

    Which really gets me nowhere because it is still unclear at to what to do with the first part of the numerator. I'm only allowed to solve this with regular limit rules, nothign fancy...
  2. jcsd
  3. Sep 22, 2008 #2


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    Hi Melawrghk! :smile:

    (have an infinity: ∞ and a square-root: √ :smile:)

    Hint: 1. can you solve limx->-∞ (√(x2+6x+9) + x)? :wink:

    2. what is (1 + x)10/9 ? :smile:
  4. Sep 22, 2008 #3
    Nope :) Because that is essentially what I'm asking. It turns out to be ∞-∞ and I'm not sure if the infinites are equal. Plus, it definitely doesn't equal -3...

    Honestly, I have no clue. 9th root of (1+x)10?

    Thanks for your help
  5. Sep 22, 2008 #4


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    Yup :biggrin:

    (what is √(x2+6x+9) ?)

    Try again! :smile:
    ok … try: what is (1 + x)10 if x is very small?

    (1 + x)5 ?

    (1 + x)10/9 ? :wink:
  6. Sep 22, 2008 #5


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    For first one try this. sqrt(x^2+6x-1)=sqrt(x^2*(1+6/x-1/x^2))=|x|sqrt(1+6/x-1/x^2). Apply that to your rationalized form.
  7. Sep 22, 2008 #6
    OOH I'm dumb! :D Okay, I get it now. Can't believe I overlooked that...
    EDIT: Wait, no. I get this one, but how would I do it with mine? Mine doesn't factor nicely...
    It's 1whatever. But if I use that tactic, won't I just end up with 0/0 again? Because if I eliminate the "h" in the numerator, I'll be left with two equal but opposite terms... Sorry, I still don't get this one.

    Thanks Dick, I'll try that.
  8. Sep 22, 2008 #7


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    Use x2 + 6x - 1 = (x2 + 6x + 9)(1 - 10/(x2 + 6x - 1)) :wink:
    (1 + x)10 = 1 +10x +15x2 + …

    so, for very small x, (1 + x)10 is approximately 1 +10x. :wink:
  9. Sep 22, 2008 #8
    Where'd you get that?...

    Yeah.. binomial expansion. But that's for a nice power. 10/9 is ANYTHING but nice.
    Maybe if I do the 9th root of 209+h to the power of 10 that would help. I'll try.

    PS. Thanks Dick, it worked :)
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