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Sigh more limit problems

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Sadly I am having more limit problems. This one involving trig.

    [itex]\lim_{x \rightarrow 0} \frac{cos2x-cosx}{x}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I am trying to work to get to terms of (1-cosx)/x because that is 0, however I am getting stuck over and over again I keep trying to do things with trig identities and double angle formulas but I am getting nowhere the answer on the calculator is 0, but I can't get to that on paper. I broke the cos2x into 2cos^2 x -1 so then I had (2cos^2x-1-cosx )/x
    breaking these up into

    [itex]\lim_{x \rightarrow a} \frac{(-1-cos2x)}{x}[/itex]-[itex]\lim_{x \rightarrow a} \frac{(1-cosx)}{x}[/itex]

    the 2nd limit will be 0 but what can I do with the first.
     
  2. jcsd
  3. Sep 25, 2007 #2
    is that (cosx)^2 or cos(2x)?
     
  4. Sep 25, 2007 #3
    cos(2x) after using the double angle formula for cos2x I used the cos^2x one to get what is up there
     
  5. Sep 25, 2007 #4
    haha i was just going to post this same problem
     
  6. Sep 26, 2007 #5
    anyone
     
  7. Sep 26, 2007 #6
    This is obviously a L'Hopital problem.
     
  8. Sep 26, 2007 #7
    no we didn't learn L'hopital's rule yet in this class
     
  9. Sep 26, 2007 #8

    Hurkyl

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    That first one doesn't exist, but that's because you made a sign error.
     
  10. Sep 26, 2007 #9

    Avodyne

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    Yes, fix the sign error. Then, if the limit of (1-cos x)/x is zero, what is the limit of (1-cos 2x)/(2x) ?
     
  11. Sep 26, 2007 #10
    I reevaluated and I got (1+cos2x)/x for the first part but this isn't (1-cos2x)/(2x)


    I had 2cos^2x so I got 2((1+cos2x)/2)/x = (1+cos2x)/(x)
     
  12. Sep 26, 2007 #11
    hospital time!

    turns out to be lim x->0 (2sin(2x)+sinx)/1

    have fun
     
  13. Sep 26, 2007 #12
    what? if your using hopital's I can't use that we didn't get there yet otherwise how did you get to that I am confused.
     
  14. Sep 26, 2007 #13
    I'm not sure where to go from (1+cos2x)/(x) for the first limit I tried to make it to sin making cos2x=2sin^2x so i had 1+2sin^x then 1-2(1-cos^x) / x which turnd to -1(1-cos^2x)/x for the first limit can someone help
     
  15. Sep 26, 2007 #14

    Avodyne

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    You still have a sign error. The formula you want to use is

    cos(2x)-cos(x) = [1-cos(x)]-[1-cos(2x)]

    Or, since you're interested in the x->0 limit, you could Taylor expand each cosine: cos(x)=1-(1/2)x2+...
     
    Last edited: Sep 26, 2007
  16. Sep 26, 2007 #15
    In the numerator of the first limit you made a sign error, the minus 1 should be 1. If you substitute x = y/2 there, then it becomes very similar to the second limit :)
     
  17. Sep 26, 2007 #16
    i don't see how I made the sign error though I used cos(2x) = 2cos^2x-1 so

    2cos^2x-1 - cosx is correct so far right?

    then I split it into two limits

    2cos^2x - 1-cos(x)

    then (2(1+cos(2x))/2)/x - 1-cos(x) ?
     
  18. Sep 26, 2007 #17

    Avodyne

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    I can't follow what you're doing. You need to write whole equations, with equal signs.

    But you're making this much more complicated than it needs to be, as should be clear from Johan's post.
     
  19. Sep 27, 2007 #18
    I understand it now thanks guys.
     
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