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!

Limits of a function containing ln an raised to a power

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    limit x --> infinity, (1+ [(ln2)/x])^x

    2. Relevant equations

    3. The attempt at a solution

    I tried raising the whole thing to e ln [function] so that the x power would get eliminated. What I've got right now is this:

    e lim(x->infinity) ln([x+ln2]/2)^x
    e lim(x->infinity) xln([x+ln2]/x)
    e lim(x->infinity) lnx + ln2
    e to the power infinity
    = infinity

    The limit is not infinity though its 2, I've done it several other ways, but it doesn't get me the right answer anytime.
  2. jcsd
  3. Sep 28, 2008 #2
    I'm not sure how you got to your third step. Since [tex]\lim_{x\to\infty}x\ln\left(1+\frac{\ln2}{x}\right)[/tex] is indeterminate, your can rewrite it in the form
    and then use L'Hopital's rule.
  4. Sep 28, 2008 #3
    Ok when I use L'Hopital's rule it goes like:

    = (x/[x+ln2])/(-1/x^2) which simplifies to -x^3/(x+ln2)

    = -3x^2

    which is still infinity
  5. Sep 28, 2008 #4
    You forgot about the chain rule.
  6. Sep 28, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Do you think that having a log in this is a problem? Would it be any easier if it were just [itex]\lim_{x\rightarrow\infty} (1+ a/x)^x ? If so, ln(2) is just a number, let a= ln(2)!
    I seem to recall (check me on this!) that
    [tex]\lim{x\rightarrow\infty} (1+ 1/x)^x= e[/itex]

    But that has no "ln(2)" or even "a" in it does it. No matter. If the problem is
    [tex]\lim{x\rightarrow\infty}(1+ a/x)^x[/tex]
    let y= x/a. Then a/x= 1/y and x= ay
    [tex](1+ a/x)^x= (1+ y)^{ay}[/tex]
    [tex]\lim_{x\rightarrow\infty}(1+ a/x)^x= \lim{y\rightarrow\infty(1+ 1/y)^{ay}[/tex]
    [tex]= \left(\lim_{y\rightarrow\infty (1+ 1/y)^y\right)^a= e^a[/tex]

    If a= ln(2) then the limit is eln(2). What is that?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Limits of a function containing ln an raised to a power