Limits and L'Hopital rule

  • Thread starter LinkMage
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  • #1
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I have to solve this:

[tex]
\lim_{\substack{s\rightarrow 0^+}} s^4 (\frac{1}{2} ln (s) - \frac{1}{8})
[/tex]

Here is what I did so far:

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

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

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

[tex]
= \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 [/tex]

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

Answers and Replies

  • #2
111
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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) :)
 
Last edited:
  • #3
shmoe
Science Advisor
Homework Helper
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Did you stop to check if the conditions of l'hopital were satisfied before you applied it a second time? What is

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

This isn't really any easier to answer than your original question.

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
18
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You mean that:

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

And that the problem ends there?
 
Last edited:
  • #5
HallsofIvy
Science Advisor
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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
18
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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
shmoe
Science Advisor
Homework Helper
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I just want to point out that at this stage:

LinkMage said:
[tex]
\lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}}[/tex]

this limit is the same as:

[tex]
\lim_{\substack{s\rightarrow 0^+}} -8s^4(\frac{1}{2} ln (s) - \frac{1}{8})^2[/tex]

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:

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