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Math Help - The sine limit

  1. #1
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    The sine limit

    I am sure you know that,
    lim (x-> 0) sin x/x = 1

    It can easily be proven by the rigorous definition of sine (either infinite series or solution to y''+y=0 with y(0)=0 and y'(0)=1).

    However, if I would have to demonstrate this to students I would like a more simple/applied/geometric demonstration.

    The one I have in my book (and another book) use the squeeze theorem. By dividing a circle sector into parts and finding the area of the circle sector. But there is a problem with this, in order to show the area of the circle is what it is we need to first show what the derivative of sin x is. And to show what the derivative of sin x is we need to know the limit of sin x/x thus it is circular (get the pun).

    I have seen this version which is much more cleaner.
    sin x < x <tan x
    On, 0<x<pi/2
    And then using the squeeze theorem appropriately.

    I can show that,
    sin x < x
    By the following diagram below (line is shortest distance between two points).
    But how do I show that,
    x < tan x

    In the diagram the angle is measure in radians and it is a unit circle. Thus, the intercepted arc is x and the blue line is sin x and the red line is tan x.
    I can show that
    blue < black (sector)
    again by saying line is shortest distance.
    Thus,
    sin x < x
    But how do I arrive at,
    black (sector) < red
    Thus,
    x < tan x?
    Attached Thumbnails Attached Thumbnails The sine limit-picture11.gif  
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  2. #2
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    Is this what you mean, PH:

    sin(x)<x<tan(x)

    sin(x)/sin(x) < x/sin(x) < tan(x)/sin(x)

    1 < x/sin(x) < sin(x)/cos(x)sin(x)=1/cos(x)

    lim_{x to 0} 1/cos(x)=1

    lim_{x to 0} x/sin(x)=1

    lim_{x to 0} sin(x)/x=1
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    Quote Originally Posted by galactus View Post
    Is this what you mean, PH:

    sin(x)<x<tan(x)

    sin(x)/sin(x) < x/sin(x) < tan(x)/sin(x)

    1 < x/sin(x) < sin(x)/cos(x)sin(x)=1/cos(x)

    lim_{x to 0} 1/cos(x)=1

    lim_{x to 0} x/sin(x)=1

    lim_{x to 0} sin(x)/x=1
    Exactly!
    But how do you get the inequality:
    sin x < x < tan x
    Geometrically?

    By inspection we can say tan x > x (in first quadrant).
    But how we justify (not I did not use the word prove) that?
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    If 0<x<Pi/2, then
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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    About 15 years ago I was assigned to teach a graduate course on proofs for in-service teachers. Of course that meant secondary teachers most of whom taught AP-Calculus. Because I had never seen a proof of this other than the one in every calculus text that I had ever seen, I set a problem for them to find different proofs. Now this was of course before the WWW but nonetheless some came up with amazing results. Unfortunately, I donít have those papers.
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    That would be interesting to see, plato.

    I just thought of another possible method.

    How about using the Taylor expansion for sinx/x and using x=0?.

    sinx/x=1-(x^2/3!)+(x^4/5!)-(x^6/7!)+(x^8/9!)-(x^10/11!)+..........

    When x=0, sinx/x=1
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    Quote Originally Posted by galactus View Post
    That would be interesting to see, plato.

    I just thought of another possible method.

    How about using the Taylor expansion for sinx/x and using x=0?.

    sinx/x=1-(x^2/3!)+(x^4/5!)-(x^6/7!)+(x^8/9!)-(x^10/11!)+..........

    When x=0, sinx/x=1
    No, you cannot do that.
    ---
    Let me explain.
    ---

    Here are two sides to defining sine and cosine:

    Mathematicians Side:
    1)We can define sine and cosine as infinite series expansions and then show the results hold.
    2)We can define sine and cosine as the unique solution guarentted by the fundamental theorem of differencial equations:
    y''+y=0 and y(0)=0 and y'(0)=1
    y''+y=0 and y(0)=1 and y'(0)=0
    Using the theory for infinite series solutions we can show that there exist an analytic solution with infinite radii of convergence (that is the sine function).

    However, there is a problem with this approach, it does not reveal any apparent application. Like for example, how is it that sine is opposite side over hypotenuse?

    Which is why I there is a geometric/intuitive approach.

    Geometric Approach:
    1)We define sine and cosine functions the standard high-school way through the unit circle.
    2)Then we prove the inequality:
    sin x < x <tan x
    Geometrically.
    3)We can show the limit of sin x/x is 1. We cannot use L'Hopital rule we still do not know what the derivative of sine is. We cannot use infinite series we do not know even if it analytic: Thus, the only reasonable way is by the inequality via squeeze theorem.
    4)We can find the derivative of sine and cosine, they are based on the fact sin/x--> 1and (1-cos x)/x---> 0 and the angle sum identities for sine and cosine.
    5)Then we show that the Lagrange remainder term approaches zero. Thus, sine and cosine are analytic functions of infinite radii of convergence.
    6)Now we know that,
    y''+y=0 for sin x and cos x on any open interval.

    Note this approach leads to the mathematicians appraoch (again mathematicians are not interested in the geometric notion of sine and cosine).

    But wait there is still more!

    7)By restricting the domain we have inverse sine and cosine functions.

    8)We can show that the area of a circle is pi*r*r by evaluating:
    4*INT_0^r sqrt(r^2 - x^2)dx
    But to evaluate this integral we need to use an inverse trigonometric function composition. We will use the substitution rule for integrals. Which is based on the derivative of sine and cosine.

    Thus, using the approach of area inequalities is "faulty" because the area formula for a circle sector is based ultimately on the limit of sin x/x.

    This is what I was thinking when I first ever seen it. Even since then I sought to show it via that inequality. But have failed. Today I was thinking about it and was very angry.

    Interestingly, I found a book which agrees with me. In the end of a chapter it says, "The approach of using area is subject to much mathematical journals, authors claim that this proof is circular and relys on the fact of the area of a circle which is taught to be true. But that does not mean it is true and thus does not constitute a proof" (Something like that, I can write the exact phrase).

    But anyway, the important thing is that there is a rigorous definition. I would be more happy if I had this intuitive approach, if I ever become a teach I will use it.
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    Quote Originally Posted by ThePerfectHacker View Post
    I am sure you know that,
    lim (x-> 0) sin x/x = 1

    It can easily be proven by the rigorous definition of sine (either infinite series or solution to y''+y=0 with y(0)=0 and y'(0)=1).

    However, if I would have to demonstrate this to students I would like a more simple/applied/geometric demonstration.

    The one I have in my book (and another book) use the squeeze theorem. By dividing a circle sector into parts and finding the area of the circle sector. But there is a problem with this, in order to show the area of the circle is what it is we need to first show what the derivative of sin x is. And to show what the derivative of sin x is we need to know the limit of sin x/x thus it is circular (get the pun).

    I have seen this version which is much more cleaner.
    sin x < x <tan x
    On, 0<x<pi/2
    And then using the squeeze theorem appropriately.

    I can show that,
    sin x < x
    By the following diagram below (line is shortest distance between two points).
    But how do I show that,
    x < tan x

    In the diagram the angle is measure in radians and it is a unit circle. Thus, the intercepted arc is x and the blue line is sin x and the red line is tan x.
    I can show that
    blue < black (sector)
    again by saying line is shortest distance.
    Thus,
    sin x < x
    But how do I arrive at,
    black (sector) < red
    Thus,
    x < tan x?
    Quote Originally Posted by ThePerfectHacker View Post
    Exactly!
    But how do you get the inequality:
    sin x < x < tan x
    Geometrically?

    By inspection we can say tan x > x (in first quadrant).
    But how we justify (not I did not use the word prove) that?
    Here's my geometric justification. Lay a string along the intercepted arc. The string has length x. Now unwind the string from the bottom. From here, the path the end of the string follows is called an involute.



    The direction the end of the string moves at every instant is perpendicular to the arc at the point being unwound. Except for at the first instant in the beginning, this direction is up and to the right. Thus the path of the end of the string as it's unwound is always above the horizontal. Where this path intersects the tangent line is the length of the string x and since the path is above the horizontal, x < tan x, which is the length of the tangent line all the way to the horizontal.
    Last edited by JakeD; December 25th 2006 at 10:46 PM.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Exactly!
    But how do you get the inequality:
    sin x < x < tan x
    Geometrically?

    ...
    Hell THP,

    if you arrange the values of sin(x), x and tan(x) a little bit different to those drawings in the previous posts, I believe the inequality is very obvious. I've attached a diagram to show you what I mean.

    EB
    Attached Thumbnails Attached Thumbnails The sine limit-sin_x_tan.gif  
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    Quote Originally Posted by earboth View Post
    Hell THP,

    if you arrange the values of sin(x), x and tan(x) a little bit different to those drawings in the previous posts, I believe the inequality is very obvious. I've attached a diagram to show you what I mean.

    EB
    Try a smaller value for x, for example, x = .1, tan(.1) = 0.100334672. Then it is not so obvious that x < tan(x).
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