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!

L'Hopitale's rule

  1. Apr 19, 2012 #1
    This is just a question. We have:

    [tex]\mathop {\lim }\limits_{x\to 0} \frac{{{x^2}\sin \frac{1}{x}}}{x}[/tex]

    Which is indeterminate of type 0/0 but we can apply L'Hopitale. Well, in fact we could (and must) cross out the x to solve the limit but let's assume that we forgot to do that and we derivate it so we get:

    [tex]\mathop {\lim }\limits_{x\to 0} \frac{{2x\sin \frac{1}{x}-\cos \frac{1}{x}}}{1}[/tex]

    And the limit doesn't exist! Why is not L'Hopitale applicable in this case not crossing out the x's?
    Thanks!
     
  2. jcsd
  3. Apr 19, 2012 #2

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Because the derivative of the numerator doesn't have a limit as x->0. The function has a limit but not the derivative. The rule does not apply if the limits of the derivatives don't exist.
     
  4. Apr 19, 2012 #3
    Hmm, well you know you can cancel the x to get: $$x \sin{\frac{1}{x}}.$$ From here it is just the sandwich theorem, we know $$ -1 \leq \sin{\frac{1}{x}} \leq 1 $$ so then $$\lim \limits_{x \to 0} -x \leq \lim \limits_{x \to 0} x \sin{\frac{1}{x}} \leq \lim \limits_{x \to 0} x$$

    And you can take it from here.
     
  5. Apr 19, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What you did wrong is doing a terrible job of differentiating the numerator. You have to use the chain rule.
     
  6. Apr 20, 2012 #5

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    No, I think the OP is correct, after simplification. (At a glance, I also thought they were wrong...but actually not).
     
  7. Apr 20, 2012 #6

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    It seemed to me the OP already knew this and referred to it. The question was why L'Hospital didn't work.
     
  8. Apr 20, 2012 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Right you are.
     
  9. Apr 20, 2012 #8
    L'hopital's rule says that whenever you have two functions f and g that each converge to zero, that if lim f'/g' exists, then lim f/g exists and the two limits are equal. It does not say that if lim f/g exists then lim f'/g' exists -- as you can see from your own example, it is entirely possible that the original limit exists and the limit of the derivatives does not.
     
  10. Apr 20, 2012 #9
    This is what I wanted to know. So then, how can I realize that f'/g' doesn't exist but f/g does (let's keep assuming I forgot to cancel the x)?

    THanks!
     
  11. Apr 20, 2012 #10
    Unfortunately I can think of no simpler way to recognize this situation than to try to compute the limit of f'/g' directly, thus showing it doesn't exist, and then to use some alternate method (such as cancelling the x and applying the squeeze theorem) to show that the limit of f/g does.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: L'Hopitale's rule
  1. Product rule (Replies: 3)

Loading...