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Prove limit of (sin2x)/(2x) as x approached 0 is 1?

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove limit of (sin2x)/(2x) as x approached 0 is 1. By prove I mean using the epsilon/delta definition of precise limit. You may use the fact that the limit of (sinx)/x as x approaches 0 is 1.


    attempt: (where E=epsilon and d=delta)

    |(sin2x)/(2x) - 1| < E if |x|<d

    2(-E+1) < (sin2x)/(2x) < 2(E+1)

    ...now im guessing that from here you need to isolate the x so as to get |x| is less than some expression, which solves for delta. But when I try this I keep getting that x is greater than some number, not less. Also I do not know what my professor means by being able to use the limit of sinx/x?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 12, 2011 #2

    micromass

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    So you know that if

    [tex]|y|<\delta~\Rightarrow~|\frac{\sin(y)}{y}-1|<\varepsilon[/tex]

    You need to find

    [tex]|x|<\delta~\Rightarrow~|\frac{\sin(2x)}{2x}-1|<\varepsilon[/tex]

    Obviously, y=2x here....
     
  4. Sep 13, 2011 #3
    I would like to know if there's any proof of "limit of (sinx)/x as x approaches 0 is 1." ..
     
  5. Sep 13, 2011 #4

    micromass

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    Sure, watch http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Sep 13, 2011 #5

    lurflurf

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    How "limit of (sinx)/x as x approaches 0 is 1." is proven depends upon how sine has been defined. If for example sin'(0)=1 in included in the definition the result is trivial.
     
  7. Sep 13, 2011 #6
    Think of a circle with radius r, and a differential angle dtheta that spans r*dtheta from the center on the circumference.

    Sin(dtheta ) = r*dtheta / r , right ( although it seems there are 2 hypotenuses both being r, take one of them accepting the other is across the angle that is about 89.999999999... degrees )

    So Sin(dtheta)*r=r*dtheta

    cancel r's

    to get to the eqn

    Sin(dtheta)/(dtheta ) = 1

    which interprets to lim x---> 0 , Sin x / x = 1
     
  8. Sep 13, 2011 #7
    Yes thanks for replies above.

    But in fact I have watched somewhere else, saying that the area of circle = pi * r^2 is dependent on the result sinx/x = 1 as x->0

    Say, cut the circle of radius r into n equal partitions (each with angle n/2pi in the center) and area of the circle = n*r^2 /2 *sin(n/2pi) (by area of triangle = A*B*1/2*sin(angle between AB)

    Take n tends to infinity, you get pi *r^2


    Is there any proofs either than geometric reasoning? Say, something like epsilon-delta stuff? thx in advance!
     
  9. Sep 13, 2011 #8

    Hootenanny

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    The most straightforward method is to use the Taylor series in the vicinity of the origin:

    [tex]\frac{\sin x}{x} = 1 - \frac{x^2}{6} +\frac{x^4}{120} + \mathcal{O}(x^6)[/tex]

    Clearly the limit is 1 as x vanishes.
     
  10. Sep 13, 2011 #9
    Oh thank you!
    But how to show that the Taylor expansion does equal to sinx without using sinx/x->1 at all? I am teaching a group of students which I want to make sure everything goes in the right path and get rid of "circular proofs"! thx!
     
  11. Sep 13, 2011 #10

    Hootenanny

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  12. Sep 13, 2011 #11
    my concern is - how can we show the area of sector is 1/2*pheta?

    and would there be any simple ways with epsilon-delta?
     
  13. Sep 13, 2011 #12

    Hootenanny

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    As I said above, you don't need to use the geometrical setting. You can simply say that [itex]\sin\theta<\theta<\tan\theta[/itex] for [itex]\theta\in(0,\pi/2)[/itex]. This could be illustrated by plotting the functions together: http://www.wolframalpha.com/input/?i=plot+sin(x),x,tan(x)+for+x=0..pi/2
     
  14. Sep 13, 2011 #13
    Thank you very much! but are there any simple proofs of x>sinx and x<tanx?
     
  15. Sep 13, 2011 #14

    lurflurf

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    sin(x) is chosen so that sin'(0)=1
    it is just a convention
     
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