Results 1 to 11 of 11

Math Help - l'H˘pital's rule for a complex function

  1. #1
    Member
    Joined
    May 2008
    Posts
    171

    l'H˘pital's rule for a complex function

    Hey guys.

    Can I use l'H˘pital's rule for a complex function?

    Lets say I want to investigate the behavior of function sin(z)/z around z=0, can I use it?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    VkL
    VkL is offline
    Member
    Joined
    Oct 2008
    Posts
    96
    What do you mean by complex? That is not a complex function.

    And You can use it in this case.

    <br /> <br />
\lim_{z\to 0} \frac{sin(z)}{z}= 1

    Since it is in undetermined form, ( 0 in the denominator and 0 in the numerator), you can take the derivative of the top and bottom, and then apply the limit again.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by asi123 View Post
    Hey guys.

    Can I use l'H˘pital's rule for a complex function?

    Lets say I want to investigate the behavior of function sin(z)/z around z=0, can I use it?

    Thanks.
    \frac{\sin(z)}{z}=\frac{e^{iz}-e^{-iz}}{2iz}=\frac{1}{2i}\left[\frac{e^{iz}-1}{z}-\frac{e^{-iz}-1}{z}\right]. A little more obvious now?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2008
    Posts
    171
    Sorry but no.
    I still cant put z=0 in that.

    I didn't quite get it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    \frac {\sin{z}} {z} = \frac {z-\frac {z^3} {3!} + ... } {z} = 1-\frac {z^2} {3!} + ...
    so the result is 1 when z \rightarrow 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    171
    Well, Thank you very much but I know the result is 1.

    My question is, can I use l'H˘pital's rule for this thing even due z is a number from the complex domain?

    I mean when you try to find limit for this thing, there are number of paths you can take in order to do that, does l'H˘pital's rule works in that case?

    Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,307
    Thanks
    1283
    Yes, L'Hopital's rule works for functions of a complex variable.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2011
    Posts
    7
    Sorry to raise a dead thread, but I have had this same question for some time and have never seen a proof of L'Hopital's Rule that does not require the mean value theorem for real differentiable functions, which as far as I have seen, does not extend to the complex numbers. Perhaps the mean value property of harmonic functions serves the same end, but I have still never come across such a proof (I must concede, I've never tried proving it myself-- it could turn out to be quite easy).

    Evidence demonstrates that L'Hopital's Rule does apply to the limit of a complex valued analytic function (as I am not able to construct a counterexample), but I would love to see a proof if anyone knows where I can find one.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    In fact, it isn't always generally true. And, if you would have googled 'L'hopitals rule complex functions' you would have found this as the fourth link. Hope it helps.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Apr 2009
    From
    MÚxico
    Posts
    721
    The proof is simple and relies on the factorization of holomorphic functions: Assume f,g: D \rightarrow \mathbb{C} are holomorphic ( D some disk around 0). If f(0)=g(0)=0 then f(z)=z\hat{f}(z) and g(z)=z\hat{g}(z) where \hat{f}(0)=f'(0) and \hat{g}(0)=g'(0), then the conclusion follows trivially. If on the other hand they both have a pole at 0 then the it's a little trickier: Take f(z)=\sum_{-k}^{\infty} b_nz^n and g(z)=\sum_{-m}^{\infty} c_nz^n be the Laurent expansions around 0 then the quotient of the functions are

    \frac{f(z)}{g(z)}= z^{m-k}\frac{b_{-k} + o(1)}{c_{-m}+o(1)}

    \frac{f'(z)}{g'(z)}=z^{m-k}\frac{-kb_{-k} +o(1)}{-mc_{-m}+o(1)}

    and in both cases the limit is equal to: 0 if m>k, \frac{b_{-k}}{c_{-m}} if m=k and \infty if m<k.

    Edit: As an afterthought notice that this proof relies on the fact that the functions are defined and not zero on a punctured neighbourhood of the limit point, so the general formulation, as in the case of a real variable, doesn't follow from this. So for example if you have one of the functions be such that doesn't extend in any way past any boundary point of a disk, then you can't use this to evaluate the limit at a boundary point.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Apr 2011
    Posts
    7
    Hi Drexel, thanks for the link. I did google it, and it actually took me here. I did not notice that link before. I am thrilled to see the conditions for the complex situation.

    Jose thank you for posting your proof.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 14th 2010, 09:56 PM
  2. Using L'Hopital's Rule :)
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 7th 2010, 09:26 PM
  3. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 28th 2009, 05:40 PM
  4. l'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 21st 2007, 08:59 AM
  5. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 2nd 2006, 09:30 AM

Search Tags


/mathhelpforum @mathhelpforum