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L'Hopital's Rule Question

  1. Sep 29, 2008 #1
    I had a test today and there was a LR limit on it that I didn't get. I thought he said he got them out of the book, but I didn't see it anywhere. The equation was:

    lim x[tex]\rightarrow[/tex][tex]\infty[/tex] of (1 + cosx(1/x))x

    So, I said that lim x[tex]\rightarrow[/tex][tex]\infty[/tex] of cos(1/x) = cos(0) = 1

    And therefore:

    lim x[tex]\rightarrow[/tex][tex]\infty[/tex] of (1 + 1)[tex]\infty[/tex] = 2[tex]\infty[/tex]

    But the only LR indeterminate for near that is 1[tex]\infty[/tex] type... I left it at that. I'm pretty sure my answer is incorrect. Anyone?
  2. jcsd
  3. Sep 29, 2008 #2
    You wrote cosx(1/x), pretty sure you meant cos(1/x)?

    Anyways, I think you are correct, take ln of both sides. You now have

    [tex] ln|y| = x \cdot ln \left(1 + cos \left( \frac{1}{x} \right) \right) = \frac{ln \left(1 + cos \left( \frac{1}{x} \right) \right)}{\frac{1}{x}} [/tex]

    If you now evaluate the limit you get

    [tex] \frac{ln(2)}{0} [/tex]

    Which as you pointed out is not in the form 0/0, note that if you had [tex] 1^{\infty} [/tex] then you would have had ln(1)/0 = 0/0
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