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Infinite limit, small question

  1. May 21, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \lim_{x \to \infty} \frac {4x}{\sqrt{2x^{2}+1}} [/tex]


    2. Relevant equations



    3. The attempt at a solution

    So at first glance I saw that this equation ends up at infinity/infinity so I tried to use L'hopital's rule and got:

    [tex] \lim_{x \to \infty} \frac {4}{2x(2x^{2}+1)^{-1/2}} [/tex]

    That seemed to have sent me in a wrong direction so I went back and just factored out the x^2 from the radical and it worked out such as:

    [tex] \lim_{x \to \infty} \frac {4x}{\sqrt{x^{2}(2+1x^{-2})}} [/tex]

    [tex] \lim_{x \to \infty} \frac {4x}{x\sqrt{2+1x^{-2}}} [/tex]

    [tex] \lim_{x \to \infty} \frac {4}{\sqrt{2+1x^{-2}}} [/tex]

    and with the limit applied:

    [tex] \lim_{x \to \infty} \frac {4}{\sqrt{2+1x^{-2}}} = \frac {4}{\sqrt{2}} = \sqrt{8}[/tex]

    which is correct, so my question is why didn't L'Hopital's Rule work when it was infinity/infinity or would it have worked out and I just stopped too soon?
     
  2. jcsd
  3. May 21, 2010 #2

    Mark44

    Staff: Mentor

    It's MUCH simpler to factor x^2 out of both terms in the radical, which is where you eventually went.
    The above can be written as
    [tex] \lim_{x \to \infty} \frac {4\sqrt{2x^2 + 1}}{2x}} [/tex]

    This is still the indeterminate form [inf/inf], and I think that repeated applications of L'Hopital's Rule won't be of any use.

     
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