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Limit of trig function

  1. Aug 28, 2014 #1
    1. The problem statement, all variables and given/known data
    lim x->0 sin4x/2x


    2. Relevant equations
    lim x->0 sinx/x =1


    3. The attempt at a solution
    can I write lim x->0 sin4x/2x as sinx/x * 4/2 = 1*2 or am I missing a step ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 28, 2014 #2

    nrqed

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    ]No, one cannot do that. There is no rule that says one can pull out a factor from a trig function like this. You have to rewrite you expression in the form
    [itex] \lim_{y \rightarrow 0} \, C \frac{\sin(y)}{y} [/itex] where C is some constant and y is something that depends on x, y=y(x). Then this limit will simply give C.

    So what you have to do is to identify what choice of y(x) will bring your limit in the form I just described.
     
  4. Aug 28, 2014 #3
    like 4/4
     
  5. Aug 28, 2014 #4

    HallsofIvy

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    I have no idea what you mean by that. But surely you know that sin(2x) is NOT equal to 2sin(x)?
     
  6. Aug 28, 2014 #5

    nrqed

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    Set aside the limit for now. Try writing the expression in the form [itex] C \, \sin(y)/y [/itex]. What is y(x)? What is C?
     
  7. Aug 28, 2014 #6
    sin4x/2x * 4/4 = 4*sin4x/4*2x
     
  8. Aug 28, 2014 #7

    LCKurtz

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    You haven't answered nrqed's questions.
     
  9. Aug 28, 2014 #8
    I'll have to get back to the forum on this
     
  10. Aug 29, 2014 #9
    C = 4,
    y(x) = 4x
     
  11. Aug 29, 2014 #10

    nrqed

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    Well, if you write [itex] C \sin(y)/y [/itex] using what you just wrote here, do you get the function that was given in the question?
     
  12. Aug 29, 2014 #11
    I believe I get sin4x/4x
     
  13. Aug 29, 2014 #12
    I don't understand this concept. Even Calculus by Larson does not explain it well.
     
  14. Aug 29, 2014 #13

    nrqed

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    Hold on. If y =4x and C = 4 then
    [itex] C \sin(y)/y = 4 \frac{ \sin(4x) }{4x} [/itex]
    right? This is not the initial expression.
     
  15. Aug 29, 2014 #14

    nrqed

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    You need the form
    [itex] C \frac{\sin(4x)}{4x} [/itex] WHat is the constant C equal to?
     
  16. Aug 29, 2014 #15

    nrqed

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    The key point is that we do not know the limit of [itex] \frac{\sin(4x)}{2x } [/itex] but we do know how to take the limit of [itex] \frac{\sin(4x)}{4x} [/itex] So you need to rewrite your initial function in the form
    [itex] \text{ constant } \times \frac{\sin(4x)}{4x} [/itex] Then you will be able to take the limit.
     
  17. Aug 29, 2014 #16
    c in the original equation was 1
     
  18. Aug 29, 2014 #17

    nrqed

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    What I meant is that we must write

    [itex] \frac{sin(4x)}{2x} = C \frac{\sin(4x)}{4x} [/itex]
    What is the value of C?
     
  19. Aug 29, 2014 #18
    It has to be 2
     
  20. Aug 29, 2014 #19

    nrqed

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    Correct.

    So you have shown that the initial function may be written as
    [itex] 2 \frac{\sin(4x)}{4x} [/itex]
    which, defining y = 4x, may be written as
    [itex] 2 \frac{\sin(y)}{y} [/itex]
    Now, when x goes to zero, y also goes to zero so you are all set to take the limit.
     
  21. Aug 29, 2014 #20
    the limit is 2.
     
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