1. ## Limit x-> infinity

Hi everyone. Got stuck again. Solved it for when x-> infinity but cant seem to get it for when x->0. Appreciate any help at all!

lim $
\frac{3^{x}+ln{(absx)}}{x^5+x^4}
$

x->0

Thanks!

2. thanks Dan! But is there a way without using that rule?

Johan

3. I thought for L'Hopital's you need both the top and bottom to tend to 0 as x tends to 0. logx tends to -infinity doesn't it?

4. You could try expanding and see what you come up with. Sometimes it makes it easier to see. Since x-->0, you shouldn't have to worry about an absolute value.

$\lim_{x\rightarrow{0}}\frac{3^{x}+ln(x)}{x^{5}+x^{ 4}}$

$=\lim_{x\rightarrow{0}}\left[\frac{ln(x)}{x+1}-\frac{ln(x)}{x}+\frac{ln(x)}{x^{2}}-\frac{ln(x)}{x^{3}}+\frac{ln(x)}{x^{4}}+\frac{3^{x }}{x+1}-\frac{3^{x}}{x}+\frac{3^{x}}{x^{2}}-\frac{3^{x}}{x^{3}}+\frac{3^{x}}{x^{4}}\right]$

5. Originally Posted by Thomas154321
I thought for L'Hopital's you need both the top and bottom to tend to 0 as x tends to 0. logx tends to -infinity doesn't it?
It can also be infinity over infinity. However that isn't the case here either. Thanks for the catch.

-Dan

6. What is the range of $ln(x)$ ? This is a question I never raised nor answered.

7. The range of $lnx$ is the domain of $e^x$, ie all real values.

8. Originally Posted by Thomas154321
The range of $lnx$ is the domain of $e^x$, ie all real values.
I suppose so; it just grows incredibly slow!