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Help with a limit

  1. Oct 8, 2004 #1
    limit as x->infinity of [(x^2-6x+1)^(1/2)-x]

    I have tried to force it into a l'hopital form without much success, and tried to look up a couple different techniques (like replacing x with 1/u and finding the limit as u->zero) but I honestly don't even know where to begin.
  2. jcsd
  3. Oct 8, 2004 #2


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    [tex] \lim_{x \rightarrow \infty} \sqrt{x^2-6x+1} - x [/tex]

    First off L'Hospital rule only works if you get when evaluating your limit

    [tex] \frac{\infty}{\infty}[/tex] or [tex] \frac{0}{0}[/tex]

    Your limit evaluates to [tex] \infty - \infty [/tex]

    To solve it multiply the expression by [tex] \frac{\sqrt{x^2-6x+1} + x}{\sqrt{x^2-6x+1} + x} [/tex]
    Last edited: Oct 8, 2004
  4. Oct 8, 2004 #3
    And following from that, you get the expression

    -6x + 1 / ((x^2-6x+1)+x)
    divide the numerator into two fractions and understand that 1 / any value of x will go to zero eventually

    also when you consider 6x watch how x^2 will increase so fast that it outpaces 6x +1 by leaps and bounds and you're left with simply x + x
  5. Oct 8, 2004 #4
    Got it, thanks =)
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