# Ln limit

## Homework Statement

find the following limit: ln((2x)^(2x))/(3x²+1) with x>infinity.

## 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?

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Mark44
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## Homework Statement

find the following limit: ln((2x)^(2x))/(3x²+1) with x>infinity.

## 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
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.
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?

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

tiny-tim