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Math Help - solving a tricky infinite limit analytically

  1. #1
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    solving a tricky infinite limit analytically

    Hi,
    I've been trying to solve this limit analytically, but I can't figure out how to manipulate the expression in such a way so that I can plug in infinity for n without getting an indeterminate form:

    lim n * sqrt(2-2cos(2pi/n))
    n ->infinity

    From entering this limit into the calculator, I get 2pi, which I know is the correct answer. So how might I solve this limit analytically?
    Also, bear in mind that this is in the precalc section, so I'm familiar only with with limit properties and such.

    Thanks a lot!
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  2. #2
    Senior Member nikhil's Avatar
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    Lightbulb

    1) 2-2cos(2pi/n)=4(sin(pi/n)^2)
    2) sqrt[2-2cos(2pi/n)]=2mod(sin(pi/n))
    now
    3) Lt n*2mod(sin(pi/n))=2pi[(sin(pi/n)]/(pi/n)] (limit n approaches to inf)
    n->inf
    but since
    4) [sinx]/x=1 if x->inf therfor
    5) Lt(2pi)[(sin(pi/n))/(pi/n)]=2pi
    n->inf
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  3. #3
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    Hi nikhil,
    Thanks a lot for your help, but you're going to fast for me! I'm unsure as to how you got that 2 - 2cos (2pi/n) = 4(sin(pi/n))^2. In addition, I'm unfamiliar with the mod notation. Can you please explain each step, and is there a way to solve the limit without using mod?
    Thanks very much,
    -Synergy777
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb

    1) cos2x=1-2(sinx)^2 its an identity
    2) |x|(mod x)=x if x>0 and -x if x<0
    now sqrt(x^2)=|x|
    but since sin(pi/n) >0 therfor |sin(pi/n)|=sin(pi/n)
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  5. #5
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    Okay, I now understand the first part, and I get 2 - 2 cos(2pi/n) = 4(sin(pi/n))^2, and after n*sqrt(4(sin(pi/n))^2), I get 2n*sin(pi/n), assuming that everything is positive since sin(pi/n) > 0. But how did you then get 2pi from the expression we now have:
    lim 2n*sin(pi/n)
    n->inf

    Also, later on you say that sin x/x = 1 if x -> inf, but isn't that = 0, not 1? And also, I know that lim of sin (n)/ n as n->0 is 0, but I'm unfamiliar with the identity that sin(pi/n)/(pi/n) as n -> inf is 1.
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  6. #6
    Senior Member nikhil's Avatar
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    Lightbulb

    sorry i did a mistake in hurry
    sinx/x=1 if x->0
    now
    consider [sin(1/n)]/(1/n) where n->inf
    let 1/n=t
    if n->inf then t->0
    so [sin(1/n)]/(1/n) where n->inf becomes
    sin(t)/t where t->0
    which is =1
    in my steps
    [sin(pi/n)/(pi/n)] n->inf
    now let pi/n=t
    if n->inf then t->0 therfor
    [sin(pi/n)/(pi/n)] n->inf =sin(t)/t where t->0
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  7. #7
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    Ah, ok, that makes sense. But, how did you figure out to divide sin (pi/n) by (pi/n) in the step:

    2n sin(pi/n) = 2pi (sin pi/n)/(pi/n)

    In other words, had we not known beforehand that the answer to this limit was 2pi, how would we know to divide by pi/n ?
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  8. #8
    Senior Member nikhil's Avatar
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    Lightbulb no need to know solution beforehand

    we know that sinx/x=1 as x->0 or
    [sin(1/n)]/(1/n)=1 as n->inf
    in the problem we had
    n*sin(pi/n)
    =[sin(pi/n)]/(1/n)
    but we need the form sinx/x (angle and denominator both are x)
    so to make angle and denominator same we multiply and divide numerator and denominator by pi so know we have
    [sin(pi/n)]/(1/n)=pi*[sin(pi/n)]/(pi/n) [it is in sinx/x form as angle and denominator are same]
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  9. #9
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    Okay, I think I got it. Thanks a lot for your help!
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