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Math Help - limit

  1. #1
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    limit

    the question is

    lim x->0 ( 3cos(2x) - sin(3x)/5x )

    however, this don't meet "0/0"
    comes out as 3(1) - 0 = 3

    how should i solve this problem?
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  2. #2
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    Hello,

    Well it's in the form constant/0, which tends to infinity (the signs depends... Since the numerator goes to 3, the sign depends on what happens in the denominator, namely +infinity if x goes to 0 from the left, and -infinity if x goes to 0 from the right)
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  3. #3
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    so do i have to go with
    x goes to 0 from the left/right??
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  4. #4
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    = lim x->0 3cos2x - lim x->0 sin3x/5x

    = lim x->0 3cos2x - 3/5 lim x->0 sin3x/3x

    what should i do next?
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  5. #5
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    where am i doing it wrong?
    can someone put me on the right track pleaes?!
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  6. #6
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    Quote Originally Posted by haebinpark View Post
    the question is

    lim x->0 ( 3cos(2x) - sin(3x)/5x )

    however, this don't meet "0/0"
    comes out as 3(1) - 0 = 3

    how should i solve this problem?
    First, please write your problem correctly- you first appeared to be talking about (3 cos(2x)- sin(3x))/(5x), not (3cos(2x)- sin(3x)/5x) as your wrote.

    If it is in fact the first of those, then, because the numerator goes to 3 and the denominator goes to 0, the limit does not exist.

    But in your last post, you were back to 3 cos(2x)- sin(3x)/(5x). If that is correct then, yes, you can break it into two limits : \lim_{x\to 0} 3 cos(2x)- \lim_{x\to 0} (3/5) sin(3x)/(3x).

    The first, \lim_{x\to 0} 3 cos(x) should be easy- cosine is continuous for all x. For you have, cleverly, multiplied both numerator and denominator by 3 to get the denominator the same as the argument of sine. I presume you did that because you know that \lim_{a\to 0}\frac{sin(a)}{a}= 1. Use that!
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