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Find out the limit of the following function

  1. Apr 12, 2012 #1
    f(0,∞)->R
    f(x)=2 [x^(sin(2x)] cos(2x)
    find lim(x->0)f(x)=?

    I have done all my hits all failed!!!
    can you please tell me how to solve it???
     
  2. jcsd
  3. Apr 12, 2012 #2

    HallsofIvy

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    Well, tell us what you have tried so we will know what hints will help.
     
  4. Apr 12, 2012 #3
    i have reached to this answer!!! is it correct..

    let y = [x^(sin(2x)] cos(2x)
    ln(y)= sin(2x)*ln(x)+ln(cos(2x))
    lim(x->0) sin(2x)*ln(x)+ln(cos(2x))
    lim(x->0) sin(2x)*ln(x) //(as ln(cos(0))=ln(1)=0
    lim(x->0)2sin(x)ln(x)
    lim(x->0)2 ln(x)/cosec(x)
    L Hopital
    lim(x->0)-2/(xcosec(x)cot(x))
    lim(x->0)-2sin2(x)/xcos(x)
    lim(x->0)-2sin2(x)/x
    once again L hopital
    lim(x->0)-4sin(x)cos(x)+2sin3(x)
    =0.................(area all steps correct)
    ln(y)=0
    y=1
    ;lim(x->0)f(x)=2y=2
    is it correct. I doesn't have marking scheme or any thing like this so i can't tell what's correct answer..
     
    Last edited: Apr 12, 2012
  5. Apr 12, 2012 #4

    SammyS

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    The correct answer is 2.
     
  6. Apr 12, 2012 #5
    you mean my answer is correct???????
    OR
    you have already done this question and saying me that answer is 2??
     
  7. Apr 12, 2012 #6

    SammyS

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    The answer is correct.

    The step that goes from
    lim(x->0) sin(2x)*ln(x)​
    to
    lim(x->0)2sin(x)ln(x)​
    is not correct, if you mean that sin(2x)ln(x) = 2sin(x)ln(x) .
     
  8. Apr 12, 2012 #7

    Dick

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    I don't think that's right.
     
  9. Apr 16, 2012 #8
    are nahi yar!!
    no i have just skipped a step..
    lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)
    cos(0)=1
    so
    lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)=lim(x->0)2sin(x)ln(x)

    so what do you think about it..
     
  10. Apr 16, 2012 #9

    Dick

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    Ok so far. I'd double check if you still think the limit is 2.
     
  11. Apr 16, 2012 #10

    SammyS

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    I'm pretty sure that [itex]\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.[/itex]
     
  12. Apr 16, 2012 #11

    Dick

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    Yeah, you're right. I've been missing the initial '2' somehow. Sorry.
     
  13. Apr 16, 2012 #12
    :smile:thanks to all of you for helping me...:smile:
     
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