1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits and L'Hopital rule

  1. Nov 25, 2005 #1
    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: Nov 26, 2005
  2. jcsd
  3. Nov 25, 2005 #2
    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
  4. Nov 25, 2005 #3

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  5. Nov 26, 2005 #4
    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: Nov 26, 2005
  6. Nov 26, 2005 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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."
     
  7. Nov 26, 2005 #6
    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?
     
  8. Nov 26, 2005 #7

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    I just want to point out that at this stage:

    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: Nov 26, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limits and L'Hopital rule
Loading...