# Limits and L'Hopital rule

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?

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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) :)

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shmoe
Homework Helper
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.

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?

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HallsofIvy
Homework Helper
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."

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?

shmoe
Homework Helper
I just want to point out that at this stage:

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

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!

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