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Math Help - limits

  1. #1
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    limits and continuity

    1. Show lim Θ/sinΘ=1
    Θ->0

    (Hint: show sinΘcosΘ <Θ<tanΘ)

    I have no idea to do this problem, other than use the squeeze theorem, but i still dont know what to do. any help would be greatly appreciated.
    Last edited by johntuan; October 4th 2008 at 06:45 AM.
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  2. #2
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    here is a link to the classical geometric approach in determining this limit ...

    http://www.csun.edu/~ac53971/courses/math350/xtra_sine.pdf
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by johntuan View Post
    1. Show lim Θ/sinΘ=1
    Θ->0

    (Hint: show sinΘcosΘ <Θ<tanΘ)

    I have no idea to do this problem, other than use the squeeze theorem, but i still dont know what to do. any help would be greatly appreciated.
    I will show my own proof for this.
    Proof.
    First draw a unit circle, and put a equilateral n-polygon in it.
    Then from the center of the circle, draw lines to the corners of the polygon.
    We see that the center angle 2\pi is divided to n, thus all the triangles have \frac{2\pi}{n} as the vertex of the isosceles triangle, and each triangles have the area \frac{1}{2}\sin\bigg(\frac{2\pi}{n}\bigg).
    Hence the sum of the areas of the triangles is A(n):=\frac{n}{2}\sin\bigg(\frac{2\pi}{n}\bigg).
    We can see that letting n tend to infinity A(n) tends to the are a of the circle \pi.
    That is
    \lim\limits_{n\to\infty}A(n)=\pi
    or equivalently
    \lim\limits_{n\to\infty}\frac{n}{2\pi}\sin\bigg(\f  rac{2\pi}{n}\bigg)=1.
    Substitute m=\frac{2\pi}{n}, then we see that m\to0 as n\to\infty, which yields
    \lim\limits_{m\to0}\frac{\sin(m)}{m}=1. \rule{0.3cm}{0.3cm}
    I hope this is helpful.
    Note. Here, m travels through rational numbers and since \lim means accumulation, we may think that m travels through reals since it is the closure of rational numbers.
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