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Square root asymptote

  1. Jun 29, 2008 #1
    Hi.

    I'd like to show that sqrt(x*(x-1)) has the asymptote x-0.5. The coefficient on "x" is found by saying

    [tex]\lim_{x\rightarrow\infty}\frac{\sqrt{x(x-1)}}{x}=1[/tex]

    but how does one find the 0.5 constant?
     
  2. jcsd
  3. Jun 29, 2008 #2

    Mute

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    [tex]\sqrt{x(x-1)} = \sqrt{x^2\left(1-\frac{1}{x}\right)} \rightarrow_{x \rightarrow \infty} x\left(1-\frac{1}{2x}\right) = x - \frac{1}{2}[/tex]

    At the arrow I've used a binomial expansion ([itex]\sqrt{1+a} \simeq 1 + a/2[/itex], to first order, when [itex]a \ll 1[/itex]) and kept the first order term, which gives the result you're looking for. (any further terms in the expansion will be [itex]\mathcal{O}(1/x)[/itex], which is why they get neglected but the 1/2 is kept, I guess.)
     
    Last edited: Jun 29, 2008
  4. Jun 29, 2008 #3
    Oh. Right. Thank you :-)
     
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