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Hard limit

  1. Feb 15, 2007 #1

    [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


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    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

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

  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

    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]

    [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

    one method

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


    (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)


    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

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    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

    [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.
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