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

One more limit with trig

  1. Feb 4, 2004 #1
    Stuck here too:

    lim as x -> 0 of [cotPxsinx]/2secx
    *P = pi

    *Thanks for your help
    Last edited: Feb 4, 2004
  2. jcsd
  3. Feb 4, 2004 #2
    This is equivalent to

    [tex]\lim_{x\rightarrow 0} = \frac{sinx}{2sin\pi x} [/tex]
    Last edited: Feb 4, 2004
  4. Feb 4, 2004 #3
    How did you get to that?
  5. Feb 4, 2004 #4
    so therefore u have
    [tex]\lim_{x\rightarrow 0} = \frac{sinx}{2sin\pi x} [/tex]
    + Using these
    cosx->1 as x->0
  6. Feb 5, 2004 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    In more detail:

    [tex] cot(\pi x)= \frac{cos(\pi x)}{sin(\pi x)}[/tex] and
    sec(x)= [tex]\frac{1}{cos(x)}[/tex]

    [tex]\frac{cot(\pi x)sin(x)}{sec(x)}= \frac{cos(\pi x)sin(x)}{sin(\pi x)cos(x)} [/tex]
    [tex]= \frac{cos(\pi x)}{cos(x)}\frac{sin(x)}{sin(\pi x)}[/tex]

    Since both [tex]cos(\pi x)[/tex] and cos(x) have limit 1 as x-> 0, the limit of [tex]\frac{cos(\pi x)}{cos(x)}= 1[/tex] and we are left with
    [tex]limit_{x->0}\frac{sin(x)}{sin(\pi x)}[/tex].

    You could do that by L'hopital's rule or by considering the first few terms of the Taylor's series for sine.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: One more limit with trig
  1. One more limit (Replies: 1)

  2. Trig limit (Replies: 2)