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!

Hard limit

  1. Feb 15, 2007 #1
    Calculate:

    [tex]lim_{z\to 0}(\frac{z}{sin z})^{\frac{1}{z^2}[/tex]

    [tex]z\in \mathbb{C}[/tex]

    I tried using Laurent series expansion but it gets too messy.
    I hope LaTex turns out properly.
     
  2. jcsd
  3. Feb 15, 2007 #2
    The bit inside the brackets approaches 1 and the 1/z^2 approaches infinity.

    Overall, would it approach 1?
     
  4. Feb 15, 2007 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    How about using the natural logarithm function ?
     
  5. Feb 15, 2007 #4
    Your idea:
    was very good.

    Simply apply it to the expression within the bracket, see what happens and think to the definition of the exponential as a limit.

    Remember: lim (1+1/n)^(x*n) = exp(x) for n->infinity
     
    Last edited: Feb 15, 2007
  6. Feb 15, 2007 #5
    {

    No.That's a typical error in thinking when you consider such limits.


    Dextercioby,how to use natural logarithm function ?
    I tried that too by taking the logarithm of the expression:

    Can you calculate:

    [tex]lim_{z\to 0}\frac{1}{z^2}*ln \frac{z}{sin z}[/tex]

    What result do you get?
     
  7. Feb 15, 2007 #6
    zoki85,

    develop the Ln(z/sin(z)) in series for small z, and you will get the result quickly,

    michel
     
  8. Feb 15, 2007 #7
    [tex]\lim_{z\to 0}\left(\frac{z}{sin\ z}\right)^{\frac{1}{Z^2}}=e^{1/6}[/tex]


    Laurent series not necessary :smile:
     
  9. Feb 16, 2007 #8
    Will you explain how to calculate the limit without Laurent please?
    I used the Laurent,but I think I got different result than your .:mad:

    lalbatros what do you get?
     
  10. Feb 16, 2007 #9
    Easily!

    Becouse of :

    [tex]\lim_{\phi\to 0}(1+\phi)^{1/\phi}=e[/tex]

    We have:

    [tex]\lim_{z\to 0}\left(\frac{z}{sin(z)}\right)^{1/z^2}=1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)=e^{1/6}[/tex]


    Limit:
    [tex]\lim_{z\to 0}\frac{sin(z)-z}{z^3}=-1/6[/tex]
    is easily obtainable if you know to apply L'Hospital rule ,and derivations for elementar holomorphic functions.
     
    Last edited: Feb 16, 2007
  11. Feb 16, 2007 #10
    zoki85,

    one method

    z/sin(z) = z/(z-z³/6) = 1/(1-z²/6) = 1+z²/6 (up to 3th order)

    therefore

    (z/sin(z))^(1/z²) = (1+z²/6)^(1/z²) = (1+u)^(1/6/u) (by a change of variable with u->infinity)

    and you get the result from tehno.

    second method

    Ln(z/sin(z)) = z²/6 (from the same series development)

    therefore

    ln((z/sin(z))^(1/z²)) = ln(z/sin(z)) * (1/z²) = 1/6

    and the same result follows ...
     
    Last edited: Feb 17, 2007
  12. Feb 20, 2007 #11
    lalbatros and tehno:Thanks a lot ,3 solutions for 1 problem !
    I had exam today (Complex Analysis) and guess what.One problem was to find:
    [tex]lim_{z\to 0}\left(\frac{sinz}{z}\right)^{\frac{sinz}{z-sinz}}[/tex]

    Your instructions were very helpful and I solved it (in two ways) with no problems.
    Thanks once again.
     
    Last edited: Feb 20, 2007
  13. Feb 20, 2007 #12

    Gib Z

    User Avatar
    Homework Helper

    Nice work :) btw if you want the z>0 to be under the lim, rather than next to it, write \lim instead of just lim in your tex.
     
  14. Feb 20, 2007 #13
    shouldn't it be
    [tex]\lim_{z\to 0}\left(\frac{z}{sin(z)}\right)^{1/z^2}=1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{1}{z^2}\ln\left(\frac{\sin z}{z}\right)}\right) = 1/\left(e^{\lim_{z\to 0}\frac{\ln \sin z - \ln z}{z^2}\right)[/tex]

    or am i missing something here?
     
    Last edited: Feb 20, 2007
  15. Feb 21, 2007 #14
    I think you don't see there's no "ln" once "e" is introduced in calculations.
    It is just an application of the definition of "e" :
    [tex]lim_{x\to 0}(1+x)^{\frac{1}{x}[/tex]
    to the function expression.
    By the way,it took me while to see that too!
     
  16. Feb 22, 2007 #15
    i still don't see exactly how tehno got:
    [tex]1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)[/tex]
     
    Last edited: Feb 22, 2007
  17. Feb 22, 2007 #16
    Look:

    [tex]\frac{sin\ z}{z}=1+\frac{sin(z)-z}{z}[/tex]

    [tex]\lim_{z\to 0}\left(\frac{sinz}{z}\right)^{1/z^2}=\lim_{z\to 0}\left(1+\left(\frac{sin(z)-z}{z}\right)\right)^{\left(\frac{z}{sin(z)-z}\right)\cdot \frac{sin(z)-z}{z^3}}[/tex]

    Do you see e now?
     
  18. Feb 23, 2007 #17
    yeah, i see it now. thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Hard limit
  1. Limit of (Replies: 13)

  2. Limit of this (Replies: 21)

  3. No limit (Replies: 3)

  4. Limit ? (Replies: 2)

  5. Hard Integrals (Replies: 32)

Loading...