Hey guys.

Can I use l'Hôpital'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.

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- Apr 5th 2010, 03:04 PMasi123l'Hôpital's rule for a complex function
Hey guys.

Can I use l'Hôpital'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. - Apr 5th 2010, 03:39 PMVkL
What do you mean by complex? That is not a complex function.

And You can use it in this case.

$\displaystyle

\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. - Apr 5th 2010, 03:44 PMDrexel28
- Apr 5th 2010, 03:49 PMasi123
Sorry but no.

I still cant put z=0 in that.

I didn't quite get it. - Apr 5th 2010, 06:20 PMxxp9
$\displaystyle \frac {\sin{z}} {z} = \frac {z-\frac {z^3} {3!} + ... } {z} = 1-\frac {z^2} {3!} + ...$

so the result is 1 when $\displaystyle z \rightarrow 0 $ - Apr 5th 2010, 11:37 PMasi123
Well, Thank you very much but I know the result is 1.

My question is, can I use l'Hôpital'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'Hôpital's rule works in that case?

Thanks a lot. - Apr 6th 2010, 04:24 AMHallsofIvy
Yes, L'Hopital's rule works for functions of a complex variable.

- May 12th 2011, 11:10 PMgosuman
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. - May 13th 2011, 06:25 PMDrexel28
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.

- May 14th 2011, 09:11 PMJose27
The proof is simple and relies on the factorization of holomorphic functions: Assume $\displaystyle f,g: D \rightarrow \mathbb{C}$ are holomorphic ($\displaystyle D$ some disk around $\displaystyle 0$). If $\displaystyle f(0)=g(0)=0$ then $\displaystyle f(z)=z\hat{f}(z)$ and $\displaystyle g(z)=z\hat{g}(z)$ where $\displaystyle \hat{f}(0)=f'(0)$ and $\displaystyle \hat{g}(0)=g'(0)$, then the conclusion follows trivially. If on the other hand they both have a pole at $\displaystyle 0$ then the it's a little trickier: Take $\displaystyle f(z)=\sum_{-k}^{\infty} b_nz^n$ and $\displaystyle g(z)=\sum_{-m}^{\infty} c_nz^n$ be the Laurent expansions around $\displaystyle 0$ then the quotient of the functions are

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

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

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

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. - May 14th 2011, 10:34 PMgosuman
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.