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Homework Help: 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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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.
     
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