Limits again

aquitaine · May 20, 2008

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 · May 20, 2008

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 · May 20, 2008

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 · May 20, 2008

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

rocomath · May 20, 2008

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

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

[tex]\lim_{x\rightarrow-\infty}\frac{\frac{e^x}{1+e^x}}{e^x}[/tex]

[tex]\lim_{x\rightarrow-\infty}\frac{1}{1+e^x}[/tex]

Answer ...

aquitaine · May 20, 2008

Right, thanks.

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