# Ln limit

1. Oct 11, 2009

### eric123

1. The problem statement, all variables and given/known data
find the following limit: ln((2x)^(2x))/(3x²+1) with x>infinity.

3. The attempt at a solution
I tried the following: lim ( ln((2x)^(2x))/(3x²+1) ) with x>infity =
lim ( (3x²+1)/ln((2x)^(2x)) ) with x>0
lim 2xln(2x) with x>0 = 0, therefore lim ( (3x²+1)/(2xln(2x)) with x>0 =
1/0 which is infinity, a simple plot however shows this is wrong, could someone help me?

Last edited: Oct 11, 2009
2. Oct 11, 2009

### Staff: Mentor

The step above isn't valid. There is no theorem that says that
$$\lim_{x \rightarrow \infty} \frac{a}{b}~=~\lim_{x \rightarrow 0}\frac{b}{a}$$
which is what it seems you have done. IOW, you can't just flip the numerator and denominator and change the limit variable.

As x gets large without bound, both numerator and denominator also get large without bound. The simplest way to do this, I believe, would be to use L'Hopital's Rule.

3. Oct 11, 2009

### eric123

okay, thank you.
I am, however, not allowed to use L´Hopital´s rule..
does anyone know a method besides L´Hopital´s rule?

4. Oct 11, 2009

### tiny-tim

Welcome to PF!

Hi eric123! Welcome to PF!

(try using the X2 tag just above the Reply box )

Hint: How fast does ln(x) go up, compared with x?