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Limit of arccosh x - ln x as x -> infinity

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data
    find the limit of arccoshx - ln x as x -> infinity

    2. Relevant equations
    ##arccosh x = \ln (x +\sqrt[]{x^2-1} )##

    3. The attempt at a solution
    ## \lim_{x \to \infty }(\ln (x + \sqrt{x^2-1} ) - \ln (x)) = \lim_{x \to \infty} \ln (\frac{x+\sqrt{x^2-1}}{x})

    \ln (1 + \lim_{x \to \infty}\frac{\sqrt{x^2-1}}{x})##

    I can see that the limit of the second part is also going to one, but I can't manipulate the expression to show this.
  2. jcsd
  3. Mar 29, 2016 #2


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    Gold Member

    I think it is ##\lim_{x\rightarrow +\infty} \ln{(x+\sqrt{x^{2}-1})}-\ln{x}=\lim_{x\rightarrow +\infty} \ln{\frac{x+\sqrt{x^{2}-1}}{x}}## that is ##\lim_{x\rightarrow +\infty} \ln{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)}##
  4. Mar 29, 2016 #3
    I was just about to come here and say "Wow I'm really stupid, I got it now". But you'd already answered. Thanks :D.
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