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

Limit of differentiable functions question

  1. Mar 13, 2009 #1
    f(x) and g(x) are differentiable on 0
    \lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2

    because i can use (sin x)/x=1 here
    is it ok??
  2. jcsd
  3. Mar 13, 2009 #2


    User Avatar
    Science Advisor

    No, you don't have "sin(x)/x", you have
    [tex]\frac{sin(\frac{f(x)+ g(x)}{2})}{x^2}[/itex]

    whether this converges at all depends on how fast f(x) and g(x) converge to 0.
  4. Mar 13, 2009 #3
    i can split x^2 into xx
    and i have two sinuses ,one for each x
    so i do have it
    the only thing that bothers me is
    its sin x/x
    the stuff inside sinus doesnt equal x
    but they both go to zero
  5. Mar 13, 2009 #4
    Well, it looks rather simple to me, but i might be totally wrong as well. Anyhow, i hope u won't kill me for this :yuck:

    I can use the fact that both f(x) and g(x) are differentiable at 0...this means that they both are continuous at 0 as well. THis is good to know.

    NOw i will alsu use the fact that cos(x) is continuous everywhere.this is good to know too.


    [tex] \lim_{x\rightarrow 0}\frac{cosf(x)-cosg(x)}{x^2}[/tex]

    Applying what i just said above, we see that we get a limit of the form 0/0.

    So, applying l'hopitals rule twice, will take you home safe.

    Edit: DO you know anything about f'(0) and g'(0)??
  6. Mar 13, 2009 #5

    you are correct
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook