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Limit of logarithmic functions

  1. Dec 30, 2009 #1
    [tex]lim_{i\rightharpoonup\infty}[/tex] [tex]\frac{ln(4^{i}-1)}{ln(2^{i})}[/tex]

    If I set this up right it should go to 2, but I'm pretty rusty and every time I try to work this out I end up getting garbage or repeating behaviors that I can't do anything with... Anyone know what exactly to do with it?

    Ack, since this isn't homework I posted it here, but since it's such a basic level could someone move it to the calculus homework forum? Sorry guys :/.
    Last edited: Dec 30, 2009
  2. jcsd
  3. Dec 30, 2009 #2
    Have u tried L'Hopitals Rule?
  4. Dec 30, 2009 #3
    and it goes to 2. another way to reason about it is, that ln(4^i-1) for i large behaves similarly with ln(4^i), so it means you can replace one with the other. THen using logarithmic rules you get as a rezult 2.
  5. Dec 30, 2009 #4
    ...I must be missing something...

    This is what I've been doing:
    http://img684.imageshack.us/img684/3658/loglimj.jpg [Broken]

    After taking the derivative and simplifying it down I end up with a similar case to what I had before (lines 3 and 9).
    Last edited by a moderator: May 4, 2017
  6. Dec 31, 2009 #5
    You can't rearrange a limit problem into a product and differentiate using the product rule; you need to differentiate the numerator and denominator separately
    [tex]\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}[/tex] (when f(c) and g(c) make the form 0/0 or ∞/∞)
    Last edited: Dec 31, 2009
  7. Dec 31, 2009 #6
    Wow, I can't believe I completely forgot how to use L'Hopital's rule properly...
    Thanks guys!
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