# Limit of logarithmic functions

1. Dec 30, 2009

### rostbrot

$$lim_{i\rightharpoonup\infty}$$ $$\frac{ln(4^{i}-1)}{ln(2^{i})}$$

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?

edit:
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. Dec 30, 2009

### sutupidmath

Have u tried L'Hopitals Rule?

3. Dec 30, 2009

### sutupidmath

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.

4. Dec 30, 2009

### rostbrot

...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
5. Dec 31, 2009

### Bohrok

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
$$\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$$ (when f(c) and g(c) make the form 0/0 or ∞/∞)

Last edited: Dec 31, 2009
6. Dec 31, 2009

### rostbrot

Wow, I can't believe I completely forgot how to use L'Hopital's rule properly...
Thanks guys!

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