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Limit/ L'Hopital's question

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the Limit (as x-> -infinity) of [(7x^2)-x+11]/(4-x)

    So I found that the limit is infinity/infinity indeterminate form so I tried to use L'hopital's to solve it.
    So I took the derivative of (7x^2)-x+11 and got 14x-1
    then for the derivative of 4-x I got -1

    So the Limit as x->-infinity of 14x-1/-1 would be -infinity/-1 . So would the limit be infinity?

    Apparently this answer is wrong. ...Is there a reason L'Hopital's wouldn't work here? How can I solve this?
    Any help available would be much appreciated, thanks!
     
  2. jcsd
  3. Dec 11, 2008 #2
    The answer should be infinity, even without l'Hospital's, simply by virtue of the quadratic in the numerator. Check to you have the right equation or right way of checking the answer.
     
  4. Dec 12, 2008 #3

    HallsofIvy

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    Why use L'Hopital? Divide both numerator and denominator by x. All fractions with x in the denominator will go to 0. What is left?
     
  5. Dec 13, 2008 #4
    Okay... that does make sense. Thanks. I get so confused with Limits working with infinity I just always manage to get myself turned around. Thanks for all the help!
     
  6. Dec 13, 2008 #5
    Denominator should be -1 right ? :smile:
    unless you plug in infinity after you applied L'Hopital and divide both numerator and denominator by x.
     
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