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What is the limit of sin2x/x as x approaches 0?

  1. Feb 8, 2014 #1
    The problem statement
    What is the limit of sin2x/x as x approaches 0?


    Revelant equations
    lim sin(x)/x = 1 x-->0

    Attempt at a solution
    so sin2x/x = 2sin(2x)/2x since sin(2x)/2x = 1 2sin(2x)/2x = 2*1

    I know how to solve it this way however my teacher said you can solve it using double angle identity.

    Sin2x= 2sinxcosx

    That would be equal to lim2sinx/x * cosx/x

    2sinx/x=2 but what would cos x/x when x approaches zero.. isnt that undefined? How do you figure that out then?
     
  2. jcsd
  3. Feb 8, 2014 #2

    BruceW

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    hey, welcome to physicsforums!

    to begin with, you had sin(2x)/x and you used sin(2x) = 2 sin(x)cos(x) right? So, using this substitution, what does your equation become? (I think you just made an unlucky error in doing the substitution, so if you think it through again, you should get the right answer).
     
  4. Feb 8, 2014 #3
    That would give me lim2sinxcosx/x

    Sorry I dont understand what I am doing wrong??
     
  5. Feb 8, 2014 #4

    BruceW

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    you've got it correct now :) (there's only one x in the denominator). From here, what does cos(x) tend to, and what does sin(x)/x tend to?

    edit: when I say 'tend to' I mean, what does it approach, in the limit
     
  6. Feb 8, 2014 #5
    Limits


    I understand that cos 0 is equal to 1 however isn't the cosx over x too ?? Since sin2x/x then by substitution , wouldn't the cos x be over x too?
     
  7. Feb 8, 2014 #6

    BruceW

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    well, as you said in post 3, you have lim 2sin(x)cos(x)/x And you can use the normal rules of multiplication here, i.e. it is of the form (ab)/c so you can rearrange as b(a/c) or a(b/c) but not (a/c)(b/c)
     
  8. Feb 8, 2014 #7

    Oh I understand it now it would only be over x if it was an addition question.
     
  9. Feb 8, 2014 #8

    BruceW

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    yeah, that's right
     
  10. Feb 8, 2014 #9

    Thank you!
     
  11. Feb 8, 2014 #10

    BruceW

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    no worries!
     
  12. Feb 8, 2014 #11

    Can you see if you can answer my other question?
     
  13. Feb 8, 2014 #12

    SammyS

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    Why not restate that "other question" ?
     
  14. Feb 8, 2014 #13

    The problem is already solved. Thank you
     
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