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Problem evaluating a limit as x to infinity algebraically

  1. Feb 15, 2009 #1

    fedaykin

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    1. My problem is such:

    Find the limit of [tex]\lim_{x \rightarrow \infty} \sqrt{9x^2+x} -3x [/tex]



    2. No relevant equations

    3. I multiplied [tex] \frac{\sqrt{9x^2+x} -3x}{1} * \frac{\sqrt{9x^2+x} +3x}{\sqrt{9x^2+x} +3x} = \frac{x}{\sqrt{9x^2+x} +3x} [/tex]

    I am now quite confused as to where to go from here. My teacher will accept tabled values, but I wish to prove my answer. I would greatly appreciate any help.

    I did attempt to divide by the highest power in the denominator, but all that got me was a mess:

    [tex] \frac{1}{\frac{\sqrt{9x^2+x}}{x} +3} [/tex]

    Ok, I'm not certain this is valid, but...:

    I'll factor out an x under the radical, then attempt to simplify

    [tex] \frac{1}{\frac{\sqrt{9x^2*(1+\frac}{x}{9x^2}{x} +3} [/tex]

    Hmm... I'm having trouble with tex and that.
     
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    Office_Shredder

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    If you pull an x out of the square root you get

    [tex] \frac{x}{x ( \sqrt{9 + 1/x} + 3)} = \frac{1}{ \sqrt{9+1/x} + 3} [/tex]

    which you should be able to do
     
  4. Feb 15, 2009 #3

    fedaykin

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    Gold Member

    Thank you so very much.
     
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