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Limit (sin(4x)/sin(6x)) as x->0

  1. Aug 29, 2014 #1
    Hi all,

    I'm just beginning calculus and I'm having trouble figuring this one out.

    I need to solve this one without using l'hospital's rule.



    1. The problem statement, all variables and given/known data

    Find

    Limit (sin(4x)/sin(6x)) as x->0

    2. Relevant equations

    We know that limit (sin(x)/x) as x -> 0 equals 1



    Thanks!!! Much appreciated!
     
  2. jcsd
  3. Aug 29, 2014 #2

    nrqed

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    In order to use the limit you now, you must rewrite the expression so that you have a factor
    [itex] \frac{\sin(4x)}{4x} [/itex] right? And you need to rewrite the [itex] \frac{1}{\sin(6x)} [/itex] in what form?
     
  4. Aug 29, 2014 #3
    OH. so i'd go

    = limit (sin(4x)/1) * Limit (1/sin(6x))

    = Limit (4sin(4x)/4) * Limit (6/6sin(6x))

    = 4/6

    =2/3

    Yeah?
     
  5. Aug 29, 2014 #4

    nrqed

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    Almost! You need to get [itex] \frac{\sin(4x)}{4x} [/itex] in order to take the limit, not
    [itex] \frac{\sin(4x)}{4} [/itex]. And same thing for the sin(6x).


    So you need to correct one step. But your final answer is correct. Good job!
     
  6. Aug 29, 2014 #5
    This can be done by
    L HOSPITAL's rule
     
  7. Aug 29, 2014 #6
    Differentiate with respect to x on denomenator and numerator and you get the answer
     
  8. Aug 29, 2014 #7
    Ans is 2/3
     
  9. Aug 29, 2014 #8

    nrqed

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    The question asks to NOT use L'Hospital's rule. Which is why I suggested this approach.
     
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