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

Homework Help: 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
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook