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Difference of Squares?

  1. Apr 21, 2007 #1
    I'm reviewing material for my exams and I came across this:

    [tex]\lim _{x\rightarrow \infty }\sqrt {{x}^{2}+x+1}-\sqrt {{x}^{2}-3\,x}[/tex]

    The only explanation it gives is "By the difference of squares" the solution sheet then jumps to:

    [tex]\lim _{x\rightarrow \infty }{\frac {4\,x+1}{\sqrt {{x}^{2}+x+1}+\sqrt

    What the hell just happened there? I can solve from then on but I've no idea what's happening on this step. Also an idiot proof link would be appreciated.
  2. jcsd
  3. Apr 21, 2007 #2


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    Think of the first line as [tex]\lim _{x\rightarrow \infty }\frac{\sqrt {{x}^{2}+x+1}-\sqrt {{x}^{2}-3\,x}}{1}[/tex], then multiply top and bottom of the fraction by [itex]\sqrt {{x}^{2}+x+1}+\sqrt {{x}^{2}-3\,x}[/itex]. Does this make the second line any clearer?
  4. Apr 21, 2007 #3
    *Hits head on wall*
    Yes, thanks.
  5. Apr 21, 2007 #4


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    I just tried it myself; how does "2" seem?
  6. Apr 21, 2007 #5


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    2 sounds good, since the function looks like
    [tex]\frac{4x}{\sqrt{x^2} + \sqrt{x^2}} = 2[/tex]
    when x is big.
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