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Math Help - please check for me!!!x goes infinity (cos(2x)-cos(3x))/(x^2)

  1. #1
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    please check for me!!!x goes infinity (cos(2x)-cos(3x))/(x^2)

    What is the limit as x goes to infinity of (cos(2x)-cos(3x)) / x^2?


    -2 <= cos(2x) - cos(3x) <= 2.

    This implies that:

    -2/x^2 <= [cos(2x) - cos(3x)]/x^2 <= 2/x^2.

    Since -2/x^2 and 2/x^2 --> 0 as x --> infinity, then, by the Squeeze Theorem, [cos(2x) - cos(3x)]/x^2 --> 0 as x --> infinity as it is bounded by -2/x^2 and 2/x^2 and the limits are the same.


    is this right??
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  2. #2
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    Quote Originally Posted by yunho520 View Post
    What is the limit as x goes to infinity of (cos(2x)-cos(3x)) / x^2?


    -2 <= cos(2x) - cos(3x) <= 2.

    This implies that:

    -2/x^2 <= [cos(2x) - cos(3x)]/x^2 <= 2/x^2.

    Since -2/x^2 and 2/x^2 --> 0 as x --> infinity, then, by the Squeeze Theorem, [cos(2x) - cos(3x)]/x^2 --> 0 as x --> infinity as it is bounded by -2/x^2 and 2/x^2 and the limits are the same.


    is this right??

    Yes it is...nice!

    Tonio
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