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Evaluate Limit

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate.


    2. Relevant equations
    [itex]lim_{x->infinity} (1 + \frac{1}{x})^x = e[/itex]

    3. The attempt at a solution

    lne = x(1+[itex]\frac{1}{x})[/itex]

    [itex]\frac{1+\frac{1}{x}}{1/x}[/itex] = 1/0 = ∞ (is this correct?
     
  2. jcsd
  3. Mar 15, 2013 #2

    Mark44

    Staff: Mentor

    No, and you're going about it the wrong way, it seems to me. You're supposed to evaluate the given limit, and show that it is e.

    One way to go is to let y = ##\lim_{x->\infty} (1 + \frac{1}{x})^x ##, then take ln of both sides. If you work with it a bit, you can get something that you can use L'Hopital's Rule on.

    BTW, in LaTeX, ∞ is \infty, and limit is \lim.
     
  4. Mar 15, 2013 #3

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Evaluate what?
    No, these are both nonsense. Surely you don't mean to claim that these equations hold for all ##x##?
     
  5. Mar 15, 2013 #4
    if i take ln i get x(1+1/x) -->> make form for L'Hopital's rule: [itex]\frac{1+(1/x)}{1/x}[/itex]

    which is 1/0 --> applying l'hopital's again just makes it more complicated but no closer to the answer needed.
     
  6. Mar 15, 2013 #5
    if i apply L'H again i get : (-1/x^2)/(-1/x^2) which would = 1 but the limit of that would be zero correct?

    if the limit = 1 then the that makes the answer e.
     
  7. Mar 16, 2013 #6

    Mark44

    Staff: Mentor

    You're leaving something out.
    If y = (1 + 1/x)x,
    then ln(y) = ln[(1 + 1/x)x]
    Now work with the right side. ln should not go away.
     
  8. Mar 16, 2013 #7

    HallsofIvy

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    Staff Emeritus
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    NO, you don't. You get x ln(1+ 1/x)

     
  9. Mar 16, 2013 #8
    sorry. well even if that is the case:

    lim (X->∞) [itex]\frac{ln(1+(1/x)}{1/x}[/itex] = 0/0


    then

    L'Hopitals -->> [itex]\displaystyle \frac{\frac{1}{1+(1/x)}}{\frac{-1}{x^2}}[/itex] = 1/0

    1/0 still keeps popping up; do i conclude that the limit of lny = ∞

    1/0 is not an indeterminate form in which l'hopital's can be applied, correct?

    furthermore, if we conclude that 1/0 = ∞

    then we have:

    lim (x->∞) lny = ∞
    lim (x->∞) [itex]e^{lny}[/itex] = [itex]e^∞[/itex]

    but i'm looking for [itex]e^1[/itex] not [itex]e^∞[/itex]
     
    Last edited: Mar 16, 2013
  10. Mar 16, 2013 #9
    i found some help on another website:

    lim_(x->oo)(1 + 1/x)^x = lim_(x->0)(1 + x)^(1/x)`

    Let `y = (1 + x)^(1/x)`

    how does lim_(x->oo)(1 + 1/x)^x = lim_(x->0)(1 + x)^(1/x)`
     
  11. Mar 16, 2013 #10
    What does the chain rule say when differentiating a function f(g(x))? Think about that and then try to re-do the step where you apply L'Hopital's rule.
     
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