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Limit question

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data
    evaluate the following limit.

    limx→∞ cos3x-cos4x/x^2 , include theorems

    2. Relevant equations

    Im guessing its a sandwhich theorem limit and that cos3x-cos4x as an upper bound of 3 and lower of -3. But was wondering if anyone can help explain to me why this is so and should I show a proof or working for this assumption.

    3. The attempt at a solution

    -3 ≤ cos 3x - cos 4x ≤ 3 , divide through by x^2 to get original equation

    -3/x^2 ≤ ( cos 3x - cos 4x ) / x2≤ 4/x2

    Now limx→∞ -3/x2 = 0 and
    limx→∞ 3/x2 = 0.

    So the limx→∞ ( cos 3x - cos 4x ) / x2 = 0 by the squeeze theorem.

    I'm not sure as to what rules I would have to put in this working out. Except maybe something

    about -1≤ cos ≤ 1 and how cos(3x) = cos(2x + x) = 3cos3-cos x

    and cos(4x) = 2 cos2(2x) - 1 = 2(2 cos2(x) - 1)2 - 1 = 8 cos4(x) - 8 cos2(x) + 1
    but i don't see how this would help.
  2. jcsd
  3. Aug 18, 2013 #2


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    limx→∞ cos3x-cos4x/x^2 does not exist. Unless you intended to write limx→∞ (cos3x-cos4x)/x^2?

    I do not see why 3 is chosen as upper bound and -3 as the lower one. But they are valid bounds. Anyway, -1≤cos3x≤1, and the same is true for cos(4x), -1≤cos4x≤1. It can happen, that cos 3x=-1 and at the same time cos 4x= 1, or cos(3x)=1 and cos(4x)=-1, So -2≤cos(3x)-cos(4x) ≤2 is always true, and you can apply the sandwich method.

    That is right, the limit is zero.

    No reason to complicate it....

  4. Aug 18, 2013 #3


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    I would use Taylor expansion of the numerator
    [tex]\cos(3 x)-\cos(4 x)=1-\frac{9 x^2}{2}-1+\frac{(4 x)^2}{2} + \mathcal{O}(x^4) = \frac{7}{2} x^2 + \mathcal{O}(x^4)[/tex]
    and then I'd think about the limit again (it's NOT 0!).

    BTW: It is good to use LaTeX for the formulae. The OP was not well defined and I could only guess what the right expression should be!
  5. Aug 18, 2013 #4


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    The Taylor expansion does not work for x→∞.

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