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Trigonometric Limit Problem

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the limit of :
    lim x-> (π/2) (2-2sin x)/(6x-3π)

    2. The attempt at a solution

    lim x-> (π/2) (2-2sin x)/(6x-3π)
    =lim x-> (π/2) 2-2 sin x / 6 (x- (1/2)pi)

    Assuming that y = x - (π/2)
    So,
    lim y->0 (2-2sin(y+pi/2))/6y
    lim y->0 (2-2 (sin y cos pi/2 + cos y sin pi/2)/6y

    Then, I substitute y=0 into the sin y cos pi/2 , so the equation remains :

    lim y-> 0 (2-2 cos y sin pi/2)/6y
    lim y->0 2(1-cos y sin pi/2) /6y
    lim y-> 0 2 (2sin^2((1/2)y) sin pi/2) /6y
    lim y->0 (4 sin^2 ((1/2)y) sin(pi/2)) /6y
    lim y->0 (2/3) (sin^2 (1/2)y) (2*(1/2)y) (sin(pi/2))

    Then, I'm stuck here, because I need one more (1/2)y as the denominator of sin(1/2)y , because sin^2(1/2)y = sin(1/2)y * sin(1/2)y, so I need two (1/2) y as the denominator
    But, I can just make one. If I make two and normalize it, it'll be zero.. But, I'm not quite sure..
    Please help me..
     
  2. jcsd
  3. Apr 23, 2015 #2
    Maybe try L'Hopital's rule?
     
  4. Apr 23, 2015 #3

    SammyS

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    Maybe not -- in the pre-calculus forum .
     
  5. Apr 23, 2015 #4

    Mark44

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    Try to keep in mind the mathematical level of the student who posts a problem. Help for a problem posted in the Precalc section should generally not use ideas or concepts from calculus. Occasionally a member will post a problem in the wrong homework section, but we mentors make an effort to move such posts to the right places.
     
  6. Apr 23, 2015 #5

    SammyS

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    No need to substitute y = 0 into sin(y) there, because cos(π/2) =0 so that sin(y)cos(π/2) =0 for all y ..

    Also, there is no need for you to drag around sin(π/2) because, sin(π/2) = 1.
     
  7. Apr 23, 2015 #6
    Aren't limits a calculus concept?
     
  8. Apr 23, 2015 #7

    SammyS

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    Sometimes yes, sometimes no. They are usually introduced before differentiation, so L'Hôpital's rule wouldn't be appropriate in that case.
     
  9. Apr 23, 2015 #8

    Mark44

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    You're right Sammy. My focus was on thinking that this problem was one in which ##\lim_{x \to 0}\frac{1 - cos(x)}{x}## was supposed to be used, and didn't notice the effect on the original problem caused by a changed sign. I've deleted that post of mine.
     
  10. Apr 23, 2015 #9

    SammyS

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    me too -- deleted it.

    It still comes to something of the form ##\displaystyle \ \lim_{y \to 0}\frac{1 - \cos(y)}{y} \ ## .
     
    Last edited: Apr 23, 2015
  11. Apr 24, 2015 #10
    Hmm.. You're right..
    Then, what should I do after :
    lim y->0 (2/3) (sin(pi/2)) (sin^2 (1/2)y)/(2*(1/2)y)
    ??

    If I make it like below
    lim y->0 (2/3) (sin(pi/2)) (sin^2(1/2)y) / ((4y)((1/2)y)^2)
    The limit will be undetermined (division by zero)
    Then, what should I do ?
     
  12. Apr 24, 2015 #11

    SammyS

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    ( You continue to keep sin(π/2) in your expressions even though sin(π/2) = 1 . )

    Have you learned that ##\displaystyle\ \lim_{x\to 0}\frac{\sin(x)}{x}=1 \ ## ?
     
  13. Apr 24, 2015 #12
    Yes, I have learned that..
    That's why I make the equation :
    lim y->0 (2/3) (sin(pi/2)) (sin^2(1/2)y) / ((4y)((1/2)y)^2)

    I notice the sin^2(1/2)y/((1/2)y)^2 becomes 1
    But, there is 4y in the denominator (to normalize the equation).. And if I plugged y = 0, it will be a division by zero
     
  14. Apr 24, 2015 #13

    SammyS

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    That would make the limit undefined. Right ?

    BUT, that 'extra' y should be in the numerator, not in the denominator.
     
  15. Apr 24, 2015 #14
    Yeah, you're right.. Got it! The limit is zero
     
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