# Limits and L'Hopital rule

1. Nov 25, 2005

I have to solve this:

$$\lim_{\substack{s\rightarrow 0^+}} s^4 (\frac{1}{2} ln (s) - \frac{1}{8})$$

Here is what I did so far:

$$\lim_{\substack{s\rightarrow 0^+}} \frac{s^4}{\frac{1}{\frac{1}{2} ln (s) - \frac{1}{8}} =$$

$$= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{\frac{-\frac{1}{2s}}{(\frac{1}{2} ln (s) - \frac{1}{8})^2}} =$$

$$= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} =$$

$$= \lim_{\substack{s\rightarrow 0^+}} \frac{12s^2}{\frac{2 (\frac{1}{2} ln (s) - \frac{1}{8})^2 + 2 (\frac{1}{2} ln (s) - \frac{1}{8})} {[2s (\frac{1}{2} ln (s) - \frac{1}{8})^2]^2}} = 0$$

Is this OK? If not, can someone help me please?

Last edited: Nov 26, 2005
2. Nov 25, 2005

### Hammie

I think you didn't get the product rule quite right in your second use of L'Hospital's.

You may find it simpler if you look at (1/2)Ln(s) to be the Ln (s^1/2)

(It MAY reduce the problem to only one use of L'Hospital's) :)

Last edited: Nov 25, 2005
3. Nov 25, 2005

### shmoe

Did you stop to check if the conditions of l'hopital were satisfied before you applied it a second time? What is

$$\lim_{\substack{s\rightarrow 0^+}}-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}$$

You might want to try going back to the start and try a different rearrangement to get it into a form you can apply l'hopital to.

4. Nov 26, 2005

You mean that:

$$\lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} = 0$$

And that the problem ends there?

Last edited: Nov 26, 2005
5. Nov 26, 2005

### HallsofIvy

Staff Emeritus
No, the problem doesn't end until you know what the limit is! (Or show that there is no limit.) Do what Shmoe said, "You might want to try going back to the start and try a different rearrangement to get it into a form you can apply l'hopital to."

6. Nov 26, 2005

Sorry, I don't understand what you mean. I arranged the function in the first step to get a "0/0" indetermination and applied L'Hopital to that.

Can you show me what you would do please?

7. Nov 26, 2005

### shmoe

I just want to point out that at this stage:

this limit is the same as:

$$\lim_{\substack{s\rightarrow 0^+}} -8s^4(\frac{1}{2} ln (s) - \frac{1}{8})^2$$

Which is more complicated than what you started with- the polynomial has the same power, and the icky term with the log has a higher power making it even ickier. This will most likely be harder to deal with than the original limit so it's a big clue you should go back to the drawing board and modify your approach.

There's more than one way to arrange what you started with to an indeterminate form you can apply l'hopital to!

Last edited: Nov 26, 2005