Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Getting derivative of the following function using L'Hopital?

  1. Dec 22, 2004 #1
    Getting limit of the following function using L'Hopital?

    I have tried this question 5 times and each time I get a diff answer. I don't know what I am doing wrong, but here's the question. Any help?

    lim
    x-> 0

    (4/x^2) - (2 / (1 - cos x))

    edited - for typo in subject heading
     
    Last edited: Dec 22, 2004
  2. jcsd
  3. Dec 22, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    First, you don't mean "getting the derivative", you mean "finding the limit".

    Second, exactly what is the function? The way you have written it, I would interpret it as (4/x^2)- 2/(1- cos x). Since the numerators of both fractions are non-zero while the denominators go to 0, there is NO limit. Certainly L'Hopital would not apply at all here.
     
  4. Dec 22, 2004 #3
    Sorry about the typo.

    Your interpretation is correct, so the answer DNE then? Furthermore, I just have a general question. How does 2sin x cosx−2x become sin 2x - 2x? I was just looking at past set solutions and it just suddenly does that. I don't understand
     
  5. Dec 22, 2004 #4
    That is due to the following trig identity:
    [tex]\sin{2x}=2\sin{x}\cos{x}[/tex]
     
  6. Dec 22, 2004 #5
    Got it. It's the double angle formula. Guess I just have to memorize it.
     
    Last edited: Dec 22, 2004
  7. Dec 22, 2004 #6
    That identity can be easily derived from the (hopefully) more familiar one below:
    [tex]\sin{a+b}=\sin{a}\cos{b}+\sin{b}\cos{a}[/tex]
    Defining both [itex]a[/itex] and [itex]b[/itex] as the the same variable [itex]x[/itex], it is obvious that we get the previous identity for [itex]\sin{2x}[/itex]. To be efficient at trigonometric limits, you should know these identities well, unless they are given to you in tests for reference.
     
  8. Dec 23, 2004 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    [tex] \lim_{x\rightarrow 0}\frac{4}{x^{2}}-\frac{2}{1-\cos x}=\lim_{x\rightarrow 0}\frac{4-4cosx-2x^{2}}{x^{2}(1-\cos x)} [/tex]

    =[tex]\lim_{x\rightarrow 0}\frac{4\sin x-4x}{2x(1-\cos x)+x^{2}\sin x}=
    4\lim_{x\rightarrow 0}\frac{\cos x-1}{2(1-\cos x)+4x\sin x+x^{2}\cos x}[/tex]


    =[tex]-4\lim_{x\rightarrow 0}\frac{\sin x}{2\sin x+4\sin x+4x\cos x+2x\cos x-x^{2}\sin x} [/tex]

    =[tex]-4\lim_{x\rightarrow 0}\frac{\cos x}{6\cos x+6\cos x-8x\sin x-x^{2}\cos x}=-\frac{4}{12}=-\frac{1}{3} [/tex]

    Daniel.
     
    Last edited: Dec 23, 2004
  9. Dec 23, 2004 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Elegant way:
    Use the fact that close to zero:
    [tex] \cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-... [/tex]
    ,then the limit becomes:
    [tex]\lim_{x\rightarrow 0}\frac{4}{x^{2}}-\frac{2}{1-1+\frac{x^{2}}{2}-\frac{x^{4}}{24}}=...=\lim_{x\rightarrow 0}\frac{48x^{2}-4x^{4}-48x^{2}}{12x^{4}-x^{6}}[/tex]
    =[tex]\lim_{x\rightarrow 0}\frac{-4}{12-x^{2}}=-\frac{1}{3} [/tex]

    Check it is still true for another term from the MacLaurin expansion of cosx
    [tex] \cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+...[/tex]

    Daniel.
     
  10. Dec 23, 2004 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Check post number 7.I guess i applied it 4 times. :wink:

    Daniel.
     
  11. Dec 23, 2004 #10
    Thanks a lot Daniel. Saw my mistake!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Getting derivative of the following function using L'Hopital?
Loading...