1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General limit formula: limit of 1/(1+2e^-x)

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Going over an old test and this question popped up and I cannot get an answer that makes sense. Somehow we were supposed to use the general limit formula to find the limit for this equation.

    2. Relevant equations

    General limit equation:
    lim f(x+h) - f(x)/h
    h→0

    3. The attempt at a solution
    nothing I do really makes sense.
     
  2. jcsd
  3. Dec 18, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Show us an attempt that doesn't seem to work. It's not really that hard. You'll wind up using l'Hopital's rule to evaluate a limit at the end.
     
  4. Dec 18, 2011 #3
    well. i was thinking of taking the ln of both sides? but then i ended up with 0. which doesnt make sense.
     
  5. Dec 18, 2011 #4
    once i applied the formula this is what i got.
    lim [(1/1+2e^-(x+h)) - (1/1+2e^-x)]/h
    h→0

    not sure how to apply l'hospitals rule to that.
     
  6. Dec 18, 2011 #5
    just tried l'hospital and got 2e^-x/(1+2e^-x)^2 - 2e^-x/(1+2e^-x)^2 = 0
     
  7. Dec 18, 2011 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Put the two terms in the numerator over a common denominator and combine them before you start doing l'Hopital. Try and factor it into parts that go to zero and parts that don't go to zero.
     
    Last edited: Dec 18, 2011
  8. Dec 18, 2011 #7
    now I've got
    lim [(2e^-x - 2e^-(x+h))/((1+2e^-(x+h))*(1+2e^-x))]/h
    h→0
     
  9. Dec 18, 2011 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok. So the 0/0 part is (2e^-x - 2e^-(x+h))/h, right? I'd simplify it a bit and use l'Hopital on it.
     
  10. Dec 18, 2011 #9
    when i apply l'hospitals rule do i just derive the 0/0 part or do i derive both parts?
     
  11. Dec 18, 2011 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The other part just approaches a nonzero number. It's a lot easier if you leave it out of the derivative and just apply l'Hopital to the 0/0 part.
     
  12. Dec 18, 2011 #11
    okay so when I derived the (2e^-x)-(2e^-(x+h))/h I got [-2e^-x + 2e^-(x+h)]/1 which ends up being zero anyway?
     
  13. Dec 18, 2011 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The derivative in l'Hopital's rule is d/dh. 2e^(-x) has no h's in it. Shouldn't it have zero derivative? It's a constant with respect to h.
     
  14. Dec 18, 2011 #13
    okay I see. so the answer is
    lim 2e^-x/9 ?
    h→0
     
  15. Dec 18, 2011 #14
    haha never mind. the answer is
    lim 2e^-x/[(1+2e^-x)^2] ?
     
  16. Dec 18, 2011 #15
    and once I have the right equation I can plug any number as x (say 1/2) and get the limit at that point? or would I have to use
    lim f(x) - f(a)/(x-a)?
    x→a
     
  17. Dec 18, 2011 #16

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It seems strange to use L'Hôpital's rule to find a limit, when that limit itself is being used to find the derivative of a function.

    However, the only other way I see to evaluate this limit is to use a Taylor series, and that is also based on derivatives.
     
  18. Dec 18, 2011 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Well, now you have the formula for general x, so sure you can just plug in the x value. And yeah, I agree with SammyS it's a little odd to use l'Hopital to find a limit when you are trying to derive a derivative. So if you can think of another way to show that limit h->0 of (1-e^(-h))/h=1 without using derivatives it would be better. On the other hand, it's a pretty complicated function to be applying the difference quotient approach to directly. So I figured a little cheating wouldn't hurt.
     
    Last edited: Dec 18, 2011
  19. Dec 18, 2011 #18
    yeah this was a question on my midterm before we had even learned L'Hospitals rule. So i don't know how they expected us to solve it haha.
     
  20. Dec 18, 2011 #19

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Then you have to use the definition of e^u. The limit is equivalent to arguing that lim u->0 (e^u-1)/u=1. The usual definition of e^u is limit n->infinity (1+u/n)^n. I think it's pretty likely that at the time you originally took the midterm they had already proved that lim u->0 (e^u-1)/u=1. I'd review what you've got in the text from that time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: General limit formula: limit of 1/(1+2e^-x)
  1. Limit problem - calc 1 (Replies: 4)

  2. Calculus 1: Limit (Replies: 4)

  3. Limit proof 1 (Replies: 2)

Loading...