# Homework Help: Violation of LHoptial

1. Apr 5, 2012

### thomas49th

Last edited: Apr 5, 2012
2. Apr 5, 2012

### clamtrox

No, the L'Hopital applies only if both g(z) and f(z) go to zero or infinity. In this case downstairs does not.

3. Apr 5, 2012

### thomas49th

so your saying I shouldn't of used L'Hopital. How else should I of solved it?

4. Apr 5, 2012

### Office_Shredder

Staff Emeritus
Simple poles in this form have a very easy way to calculate the residue. Consider f(z)/z. Suppose f(z) is holomorphic at z=0 (note f(z) might be, for example/ 1/(1+z), so you CAN write your problem in this form).

Then the Laurent series for f(z)/z is
$$f(0)/z+ f'(0) + f''(0) z /2 +....$$

just taking the Taylor series for f(z) and dividing it by z. Is it clear now what the residue is?

5. Apr 5, 2012

### D H

Staff Emeritus
L'Hôpital's rule is perfectly valid here. The function f(z) has two poles, one at i/3 and the other at -i/3. Looking at the latter (your example), we need to calculate
$$\lim_{z\rightarrow -i/3} (z+i/3)f(z) = \lim_{z\rightarrow -i/3}\frac{(z+i/3)z}{9z^2+1} = \lim_{z\rightarrow -i/3}\frac{z^2+i/3\,z}{9z^2+1}$$
At the limit we have something of the form 0/0, so L'Hopital's rule applies. Taking the derivatives of the numerator and denominator yields
$$\lim_{z\rightarrow -i/3} (z+i/3)f(z) = \lim_{z\rightarrow -i/3}\frac{2z+i/3}{18z} = \frac 1 {18}$$
At the limit, the numerator and denominator are -i/3 and -18i/3, respectively, so the residue is 1/18.

The same analysis works for the other pole as well.

6. Apr 5, 2012

### Ray Vickson

If we have a function of the form $$F(z) = \frac{f(z)}{(z-a) g(z)}$$ with $g(a) \neq 0,$ the residue of F at a is just $$\text{Res}(F)(a) = \frac{f(a)}{g(a)}.$$ There is no need to use L'Hospital's rule here, because the work performed by the rule has already been done. Here is what I mean: you could define $\text{Res}(F)(a) = \lim_{z \rightarrow a} (z-a) F(z).$ If you do that, notice that we have
$$(z-a) F(z) = (z-a) \frac{f(z)}{(z-a)g(z)} = \frac{f(z)}{g(z)},$$ which has a perfectly nice limit equal to $f(a)/g(a),$ with no need to use L'Hospital.

In fact, you are looking at the problem backwards: the whole point of L'Hospital is to "discover" the powers m and n (if any) in the expression $$\frac{N(z)}{D(z)} = \frac{(z-a)^n u(z)}{(z-a)^m v(z)}, \text{ where } v(a) \neq 0.$$ In the current problem you don't need to do that, because it has already been done for you.

RGV

7. Apr 5, 2012

### thomas49th

Ok, thanks. I understand, haven't looked at the Laurent series in detail but looks it looks ok.
While we are on the subject of limits I have got

$$Lim_{z \rightarrow ia} \frac{(z-ia)}{(z^2 + a^2)(z^2+b^2)}e^{imz}$$

Is there a quick way to find it's residue without multiplying out then taking the derivative of upstairs and downstairs until something happens?

edit: what is with this new itex. Itallic tex?

Last edited: Apr 5, 2012
8. Apr 5, 2012

### Ray Vickson

Your expression looks OK to me, but maybe you don't like $Lim,$ and would rather have $\lim$ instead? If so, write "\lim" to override the math italics (same as saying "\exp" instead of "exp", "\sin" instead of "sin", etc.)

Anyway, you can write $z^2 + a^2 = (z - ia)(z + ia),$ then cancel out the $(z - ia)$ factor in both the numerator and denominator. You can just put z = ia in what is left. So, basically, you can just write down the answer with hardly any work at all.

RGV

9. Apr 5, 2012

### thomas49th

Thanks, I didn't spot that :)