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!

Used l'hopitals rule too many times?

  1. Apr 21, 2009 #1

    a.a

    User Avatar

    1. The problem statement, all variables and given/known data
    lim (sin x - x)/x as x--> 0

    2. Relevant equations



    3. The attempt at a solution

    I used L'hopitals rule 3 times and got -1/6, but there is actually an asymptote at x = 0 so the limit should approach infinity or -infinity (in this case -infinity)
    Where did I go wrong? Is it because I used l'hopitals rule too many times or inappropriately?
     
  2. jcsd
  3. Apr 21, 2009 #2

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    Why did you use it three times? You just need to use it once.
     
  4. Apr 21, 2009 #3

    a.a

    User Avatar

    Re: Limit

    but u still get 0/0 when u do this
    sin x- x/x^3
    =cos - 1 /3x^2
    subs x=1
    =0/0
     
  5. Apr 21, 2009 #4

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    Not sure what you did there. If you differentiate the top and bottom, you get:

    (cos(x) - 1)/1
     
  6. Apr 21, 2009 #5

    a.a

    User Avatar

    Re: Limit

    sorry the original question:
    lim (sin x - x)/x^3 as x-->0
     
  7. Apr 21, 2009 #6

    a.a

    User Avatar

    Re: Limit

    so u get:
    (cos(x) - 1)/(3x^2)
     
  8. Apr 21, 2009 #7

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    Yes, in that case you use L'Hospital twice. And your answer of -1/6 is correct.
     
  9. Apr 21, 2009 #8

    a.a

    User Avatar

    Re: Limit

    But if you take x = 0.0001 you get a very small -ve number
    and take x = -0.0001 you get a small positive number, as if there is an asymptote....
     
  10. Apr 21, 2009 #9

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    It should be negative for -0.0001 too.
     
  11. Apr 21, 2009 #10

    a.a

    User Avatar

    Re: Limit

    so why do we get an actual number for the limit rather than -infinity?
     
  12. Apr 21, 2009 #11

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    Why do you think it should be minus infinity? What expression did you get after applying L'Hospital's rule twice?
     
  13. Apr 21, 2009 #12

    a.a

    User Avatar

    Re: Limit

    Because I was told that the graph has an asymptote at x=0
    I was able to solve using L'hopitals rule and did not get an asymptote. That's why I'm confused...
     
  14. Apr 21, 2009 #13

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    No there's no asymptote at x = 0. What exactly do you mean by asymptote?
     
  15. Apr 21, 2009 #14

    dx

    User Avatar
    Homework Helper
    Gold Member

  16. Apr 21, 2009 #15

    a.a

    User Avatar

    Re: Limit

    Sorry, I guess the teacher was wrong, the function is defined at x=0 then.
    Thanks
     
  17. Apr 21, 2009 #16

    Mentallic

    User Avatar
    Homework Helper

    Re: Limit

    I'm unfamiliar with L'hospital's rules of differentiation, but:

    d/dx((sinx-1)/x^3)=(xcosx-3sinx+3)/x^4

    rather than what you both did which was differentiating the numerator and denominator separately.
     
  18. Apr 21, 2009 #17

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Re: Limit

    L'hopital's rule has you differentiating the bottom and the top separately. It's not a rule of differentiation but a rule of limits, namely that if the limit of a quotient is in one of several forms (0/0 or such) then it is the same as the limit of the quotient of the derivatives. When they say "take the derivative" they mean "take the derivative of the top and then the bottom".
     
  19. Apr 21, 2009 #18

    Mentallic

    User Avatar
    Homework Helper

    Re: Limit

    Ahh thankyou Matterwave.
     
  20. Apr 22, 2009 #19

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Limit

    I don't see why you would use L'Hopital at all.
    [tex]lim_{x\rightarrow 0}\frac{sin x- x}{x}= \lim_{x\rightarrow 0}\frac{sin x}{x}- 1= 1- 1= 0[/tex]
     
  21. Apr 22, 2009 #20

    dx

    User Avatar
    Homework Helper
    Gold Member

    Re: Limit

    Hi Halls,

    He posted the question wrong, see post #5.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Used l'hopitals rule too many times?
Loading...