1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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....


    ehild
     
  4. Aug 18, 2013 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The Taylor expansion does not work for x→∞.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit question
  1. Limit Question (Replies: 6)

  2. Limit question ! (Replies: 2)

  3. Limit question (Replies: 10)

  4. Limits question (Replies: 6)

  5. Limits question (Replies: 5)

Loading...