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Math Help - Solving a trig limit

  1. #1
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    Solving a trig limit

    I've been trying to work out this limit problem for the past few days. I got a few friends to look at it, and they haven't been able to solve it either. I think I'm just overlooking something...but I'm almost positive I need to apply the cosine double angle identity...I just get stuck after that. Here's the problem:

    lim cos(2x) - cos(x)
    x->0 ----------------
    x


    I just can't figure out how to evaluate it. Any help would be appreciated.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle\lim_{x\to 0}\frac{\cos 2x-\cos x}{x}=\lim_{x\to 0}\frac{-2\sin\frac{x}{2}\sin\frac{3x}{2}}{x}=
    \displaystyle =-\lim_{x\to 0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)  \cdot\lim_{x\to 0}\sin\frac{3x}{2}=-1\cdot 0=0
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  3. #3
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    Such a fast response...wow, thanks a bunch. What identity did you use? I just want to make sure I understand what you did so I can do it later myself if I have to.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    here's an alternate method:

    \lim_{x \to 0} \frac {\cos 2x - \cos x}x = \lim_{x \to 0} \frac {2 \cos^2 x - 1 - \cos x }x

    = \lim_{x \to 0} \frac {(2 \cos x + 1)(\cos x - 1)}x

    = \lim_{x \to 0} (2 \cos x + 1) \cdot \lim_{x \to 0} \frac {\cos x - 1}x

    = 3 \cdot 0

    = 0
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    here's an alternate method:

    \lim_{x \to 0} \frac {\cos 2x - \cos x}x = \lim_{x \to 0} \frac {2 \cos^2 x - 1 - \cos x }x

    = \lim_{x \to 0} \frac {(2 \cos x + 1)(\cos x - 1)}x

    = \lim_{x \to 0} (2 \cos x + 1) \cdot \lim_{x \to 0} \frac {\cos x - 1}x

    = 3 \cdot 0

    = 0
    Hmm, that's a method I was trying out...but when I got to the part when you split up the two functions, I kept the x as the denominator for both parts. I thought this was required...but in your example, I guess this isn't necessary?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gamerdude View Post
    Hmm, that's a method I was trying out...but when I got to the part when you split up the two functions, I kept the x as the denominator for both parts. I thought this was required...but in your example, I guess this isn't necessary?
    you cannot do that. remember, \frac 1x \cdot \frac 1x = \frac 1{x^2}. so if you distributed the x among both fractions, you would be changing the value
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    you cannot do that. remember, \frac 1x \cdot \frac 1x = \frac 1{x^2}. so if you distributed the x among both fractions, you would be changing the value
    Ah, gotcha. That makes sense. I get mixed up on the simplest things sometimes...it's embarrassing. Thank you for your help, both of you. I greatly appreciate it. I thanked both of you for your helpful responses.
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