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Find the limit as x approaches infinity of (sqrt(1+3x^2))/x

  1. Feb 24, 2009 #1

    KevinL

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    1. The problem statement, all variables and given/known data
    Find the limit as x approaches infinity of (sqrt(1+3x^2))/x

    3. The attempt at a solution
    I tried using l'hopital's rule, but it gave me 3x/(sqrt(1+3x^2)) which doesnt help me at all.
     
    KevinL, Feb 24, 2009
  2. jcsd
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  3. Feb 24, 2009 #2

    rock.freak667

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    Homework Helper

    Re: limit

    use the fact that [itex]x= \sqrt{x^2}[/itex]
     
    rock.freak667, Feb 24, 2009
  4. Feb 24, 2009 #3

    jgens

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    Gold Member

    Re: limit

    Well, I can think of a couple of ways to do this one. Probably the simplest is to note that for arbitrarily large x, 3x^2 + 1 ~ 3x^2. Another way would be to multiply and divide the equation by x and then try to find the limit.
     
    jgens, Feb 24, 2009
  5. Feb 24, 2009 #4

    JG89

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    Re: limit

    Factor out a sqrt(x^2) from the numerator.
     
    JG89, Feb 24, 2009
  6. Feb 24, 2009 #5

    KevinL

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    Re: limit

    I dont believe you can just do that. Only if two things are multiplied beneath a radical can something be taken out.

    Multiplying by x/x does not help either.

    Hmm, perhaps you can say that as x gets arbitrarily large, its sqrt(3x^2)/x because the 1 becomes meaningless and thus the limit is sqrt(3)?
     
    KevinL, Feb 24, 2009
  7. Feb 24, 2009 #6

    jgens

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    Re: limit

    Yes you can do that! However, multiplying and dividing by x certainly does help: Lim (x -> infinity) sqrt(1/x^2 + 3)/1.
     
    jgens, Feb 24, 2009
  8. Feb 24, 2009 #7

    JG89

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    Re: limit



    [tex]\sqrt(1+3x^2) = \sqrt(x^2(\frac{1}{x^2} + 3)) = \sqrt(x^2) \sqrt(\frac{1}{x^2} + 3)[/tex]
     
    JG89, Feb 24, 2009
  9. Feb 24, 2009 #8

    KevinL

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    Re: limit

    Ah, clever. Thank you. Thats definitely how my prof would want me to do it.
     
    KevinL, Feb 24, 2009
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