Finding the Limit of ln(1+e^x)/e^x: A Puzzling Problem

[aquitaine]

Problem: Find the limit of ln(1+e^x)/e^x as x approaches negative infinity

I really have no idea how to do this kind of limit, so I guessed. I tried to substitute u=e^x to get ln(1+u)/u as u approaches negative infinity, then applying l'hopital's rule and eventually ended up with 1/negative infinity which is zero. This wasn't the right answer (which was 1). Appearently what I did wasn't the right thing, so what is the right way for this kind of problem?

 

[rocomath]

As x goes to infinity, we have an indeterminate form of 0/0. So apply L'Hopital's rule.

Show us your work because using L'Hopital should have worked because you messed up your algebra.

 

[aquitaine]

Ok, substituting u = e^x

lim ln(1+u)/u as u approaches negative infinity.

Applying lhopital's rule we have

1/(u+1)/1

so the limit of 1/(u+1) as u approaches negative infinity = 1/(negative infinity+1) = 1/negative infinity = 0

Which part did I mess up?

 

[daveb]

As x goes to negative infinity, what does u go to?

 

[rocomath]

Quit substituting, you don't need to at all.

\lim_{x\rightarrow-\infty}\frac{\ln{(1+e^x)}}{e^x}

\lim_{x\rightarrow-\infty}\frac{\frac{e^x}{1+e^x}}{e^x}

\lim_{x\rightarrow-\infty}\frac{1}{1+e^x}

Answer ...

 

[aquitaine]

Right, thanks.