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Basic Limit Involving x-xcosx and sin^2(3x)

  1. Sep 1, 2011 #1
    My answer is correct, but I feel like I might've taken an illegitimate route to get there. Can anyone verify? Many thanks in advance.

    1. The problem statement, all variables and given/known data
    [tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]

    2. Relevant equations
    [tex]\sin ^{2}x = \frac{1-\cos 2x}{2}[/tex]
    [tex]\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}[/tex]



    3. The attempt at a solution
    [tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]
    [tex]=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{(\cfrac{1-cos(6x)}{2})}[/tex]
    [tex]=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{1}\cdot\frac{2}{1-cos(6x)}[/tex]
    [tex]=\underset{x\rightarrow0}{\lim}2((x-x)cosx)\cdot(1-cos(6x))[/tex]
    [tex]=2((0)cos0)\cdot(1-cos0)[/tex]
    [tex]=2(0\cdot1)\cdot(1-1)[/tex]
    [tex]=0[/tex]
     
  2. jcsd
  3. Sep 1, 2011 #2

    Ray Vickson

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    The second line of (3) is wrong: you cannot write x - x*cos(x) as (x-x)*cos(x). However, you can write it as x*2*sin(x/2)^2, so your function is 2*x*[sin(x/2)/sin(3x)]^2.

    RGV
     
  4. Sep 1, 2011 #3
    Never mind I see it now:

    sin^2(3x) = sin^2(2x + x) = [cos(2x)sin(x) + sin(2x)cos(x)]^2

    which ends up giving 0/4 = 0
     
  5. Sep 2, 2011 #4
    Just for fun I solved it using your idea as well:

    [tex]\lim_{x\rightarrow 0}\frac{x-xcosx}{sin^{2}3x}[/tex]
    [tex]=\lim_{x\rightarrow 0}\frac{x(1-cosx)}{sin^23x}[/tex]
    [tex]=\lim_{x\rightarrow 0}\frac{x(2sin^2(\frac{x}{2}))}{sin^23x}[/tex]
    [tex]=\lim_{x\rightarrow 0}\frac{xsinx}{\frac{2}{sin^23x}}[/tex]
    [tex]=\lim_{x\rightarrow 0}\frac{xsinx}{2}\cdot \frac{1}{sin^23x}[/tex]
    [tex]=\lim_{x\rightarrow 0}\frac{xsinx\cdot csc3x}{2}[/tex]
    [tex]=\frac{0}{2}[/tex]
    [tex]=0[/tex]
     
  6. Sep 2, 2011 #5

    vela

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    Your last attempt has numerous errors in it, and I don't see how what you wrote in post #3 helps in evaluating the limit.
     
  7. Sep 2, 2011 #6
    Oops. Can you point me towards my errors in my last attempt?
     
  8. Sep 2, 2011 #7

    vela

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    Going from the third to fourth lines, fifth to sixth lines. Finally, when evaluating the limit at the end, you seemed to pull a 0 out of nowhere. Cosecant diverges as x goes to 0, so you have an indeterminate form.
     
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