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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
    cotpix=cospix/sinpix.
    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

    HallsofIvy

    User Avatar
    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]

    so
    [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.
     
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