Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits again

  1. May 20, 2008 #1
    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?
  2. jcsd
  3. May 20, 2008 #2
    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.
  4. May 20, 2008 #3
    Ok, substituting u = e^x

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

    Applying lhopital's rule we have


    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?
  5. May 20, 2008 #4
    As x goes to negative infinity, what does u go to?
  6. May 20, 2008 #5
    Quit substituting, you don't need to at all.




    Answer ...
    Last edited: May 20, 2008
  7. May 20, 2008 #6
    Right, thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook