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Basic Limit

  1. Apr 3, 2010 #1
    How do I calculate the limit:
    [itex]lim x-> infinity ((2x+1)^13(3x-7)^17)/((3x+5)^30) [/itex]

    Should I use the binomial theorem to open it up?
     
  2. jcsd
  3. Apr 3, 2010 #2

    jack action

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    "[URL [Broken] rule[/URL]
     
    Last edited by a moderator: May 4, 2017
  4. Apr 3, 2010 #3
    I can't use L'Hopital becuase we didn't learn it yet. That's what making this question troubling. The power is 30 in the numerator and the denominator so i'm not sure how to work around it.
     
  5. Apr 3, 2010 #4

    Office_Shredder

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    When taking the limit of a rational function as x goes to infinity you only need to consider the highest degree term in the denominator and the numerator, because the rest of the terms don't grow as fast
     
  6. Apr 3, 2010 #5
    Well that's exactly the problem, the denominator and the numerator's power is 30 so it's obviously the same degree. What do I do from this point?
     
  7. Apr 3, 2010 #6

    Office_Shredder

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    If you had to find the limit as x goes to infinity of [tex] \frac{5x^7}{12x^7}[/tex], it's obviously 5/12 right? Same deal, find what the coefficient of each term is and that's your limit
     
  8. Apr 3, 2010 #7

    jack action

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    Well, that method is a lot faster than mine!
     
  9. Apr 3, 2010 #8

    Office_Shredder

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    If you do l'hopital's rule, at the end you'll just get the ratio of the leading coefficients, with each being multiplied by n! where n was the degree of the numerator and denominator, so it's easy to see how L'hopital implies this method
     
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