Here is the proof from my Calc book. I wonder if you have any qualms about this proof. Are you okay with using the squeeze theorem?
EDIT: sorry I can't resize that on my phone. Here's the link:
http://m.imgur.com/qNX0VLY
Wanted: Elegant proof of sin(x)/x limit
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 jestingrabbit
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Re: Wanted: Elegant proof of sin(x)/x limit
first inequality needs tan(x) > x, which is one of the things that Qaanol wanted actual proof for.
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Re: Wanted: Elegant proof of sin(x)/x limit
As I recall, he was explicitly okay with saying that the triangle contains the sector, because the area inequality for convex objects has a straightforward proof. Which is all that first inequality depends on. My best guess as to his complaint would be the assertion that the area of the sector is Theta/2.
Re: Wanted: Elegant proof of sin(x)/x limit
quintopia wrote:Here is the proof from my Calc book. I wonder if you have any qualms about this proof. Are you okay with using the squeeze theorem?
I described that proof in the first post of this thread as among “the most promising ones I’ve seen”, and pointed out that “the difficult step is to show that a sector of the unit circle has area equal to half its arc length.”
Yes, the squeeze theorem is fine. If we accept that a circle has lesser perimeter than any circumscribing polygon, then we can use the squeeze theorem to prove the required ratio between area and circumference.
We could also get that ratio by an argument about the speed of a line rotating about its end in the plane: for any x in [0, 1] the two points at distance x and (1x) together have combined speed equal to the point at distance 1, so the average speed of the line is half the speed of that endpoint. Every point moves orthogonal to the line, so the total area swept out is half the line’s length times the outer arc it describes.
Both approaches rely on a certain amount of intuition. I think Archimedes’ claim about nested convex shapes is more “basic” in some sense though.
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Re: Wanted: Elegant proof of sin(x)/x limit
How about proving the asin(x)/x limit approaches 1, instead?
Consider the set of circles tangent to the Y axis at the origin, extending into +x, and radius at least 1. For each circle, let r be its radius, and let A(r) be the point closest to the y axis that the circle intersects the line y=1. The origincenterA angle = asin(1/r).
As r goes to infinity, A(r) approaches (0,1).
I think that line of reasoning might be more direct than what I've seen here.
Consider the set of circles tangent to the Y axis at the origin, extending into +x, and radius at least 1. For each circle, let r be its radius, and let A(r) be the point closest to the y axis that the circle intersects the line y=1. The origincenterA angle = asin(1/r).
As r goes to infinity, A(r) approaches (0,1).
I think that line of reasoning might be more direct than what I've seen here.
Re: Wanted: Elegant proof of sin(x)/x limit
From the definition as the side of a right triangle of unit hypotenuse:
Lemma 1: sin t<t, t>0
Proof: consider the line y=tx. Then at x=1, we get y=t. But then x^2+y^2=1+t^2>1, and this point is outside the unit circle. But we started inside the circle, and as we have a continuous path, we must cross the circle. As we have a straight line, it must do so in a triangle. But this gives us a valid value for sin(t) by drawing a line to the y axis. Thus sin(t) is earlier on the path, and as such has a smaller value of y by monotonicity. Thus sin(t)<t
Lemma 2:
For any 1>d>0, there is a U s.t. for u<U, sin(u)>(1d)u
Proof: Let 0<d<1. Then (1d) is also in (0,1). Now, consider the line x=(1d). Then the point (1d,0) is inside the circle, and (1d,1) is outside the circle. Then as this line is a continuous one, it crosses the unit circle at some point, say (1d,b). We also show this can happen only once : if (1d)^2 +y^2=>1, e>0, then (1d)^2+(y+e)^2=(1d)^2+y^2+2ey+e^2>1. Thus if for some y it is out of or on of the circle, it is outside for all greater y. Thus it is inside the circle for all y<b. Now for u<U=b/(1d), consider the line y=ux. Then as (1d,u(1d)) has a y value of u(1d)<U(1d)=b, we have this pointinside the circle. But by the same argument as before, we cannot recross the circle, so must pass this point on the line before crossing the circle, and as our slope is positive, we must thus have (1d)u<sin u
Theorem 3: sin(x)/x>1 as x>0
Proof: we have sin(x)/x <1 for x>0 as sin(x)<x. Also by lemma 2, d=x, sin(x)>(1x)x ==>sin(x)/x >1x. But now we have squeezed sinc(x) between two functions tending to 1 for x>0. We have sin(x) = sin(x) by reflection, so sin(x)/(x) = sin(x)/x, and we have validity for x<0 .
Typed in bed, so let me know if anything is wrong/needs clarifying. And this seemed a lot more elegant before realising how many obvious things I had to prove! Perhaps this is a case of a icture speaking 1000 words...
Lemma 1: sin t<t, t>0
Proof: consider the line y=tx. Then at x=1, we get y=t. But then x^2+y^2=1+t^2>1, and this point is outside the unit circle. But we started inside the circle, and as we have a continuous path, we must cross the circle. As we have a straight line, it must do so in a triangle. But this gives us a valid value for sin(t) by drawing a line to the y axis. Thus sin(t) is earlier on the path, and as such has a smaller value of y by monotonicity. Thus sin(t)<t
Lemma 2:
For any 1>d>0, there is a U s.t. for u<U, sin(u)>(1d)u
Proof: Let 0<d<1. Then (1d) is also in (0,1). Now, consider the line x=(1d). Then the point (1d,0) is inside the circle, and (1d,1) is outside the circle. Then as this line is a continuous one, it crosses the unit circle at some point, say (1d,b). We also show this can happen only once : if (1d)^2 +y^2=>1, e>0, then (1d)^2+(y+e)^2=(1d)^2+y^2+2ey+e^2>1. Thus if for some y it is out of or on of the circle, it is outside for all greater y. Thus it is inside the circle for all y<b. Now for u<U=b/(1d), consider the line y=ux. Then as (1d,u(1d)) has a y value of u(1d)<U(1d)=b, we have this pointinside the circle. But by the same argument as before, we cannot recross the circle, so must pass this point on the line before crossing the circle, and as our slope is positive, we must thus have (1d)u<sin u
Theorem 3: sin(x)/x>1 as x>0
Proof: we have sin(x)/x <1 for x>0 as sin(x)<x. Also by lemma 2, d=x, sin(x)>(1x)x ==>sin(x)/x >1x. But now we have squeezed sinc(x) between two functions tending to 1 for x>0. We have sin(x) = sin(x) by reflection, so sin(x)/(x) = sin(x)/x, and we have validity for x<0 .
Typed in bed, so let me know if anything is wrong/needs clarifying. And this seemed a lot more elegant before realising how many obvious things I had to prove! Perhaps this is a case of a icture speaking 1000 words...
Re: Wanted: Elegant proof of sin(x)/x limit
ZeroPoint wrote:we must thus have (1d)u<sin u
This is the part that you have not sufficiently demonstrated.
In particular, the line y = ux does not cross the unit circle at height sin(u), but rather at height sin(atan(u)) = u/√(1+u^{2}).
wee free kings
Re: Wanted: Elegant proof of sin(x)/x limit
Youare right  i really messed that one up.
Reproof:
We have cosx tendingto 1 as x tends to 0, from our above sin(x) <x, and sinx >=0 for x>=0 giving sinx tends to zero. Thus cos^2x tends to 1, and so does cos x(as it is certainly positive). Now take e>0 and x such that cosx>1e. We want to show (1e)x<sinx.
Proof: consider a line inclined at angle x, and concentric circles at radius 1e and 1. Then (1e)x is the arc length between the x axis and the line. Now this distance is bounded above by (1e)(1cosxe) +(1e)sinx=(1e)(sinx+(1e)cosx)<sinx + (1ecosx)<sinx. This is from drawing a rectangle on the smaller circle bounding the arc, and then the bound is clear, as if not a circle would not be the shape of maximal area by puting this rectangle into a circle.
Edit: I messed this up again  the bound should be (1e)(sinx+1cosx). Back to the drawing board...geometry clearly isnt my forte
Reproof:
We have cosx tendingto 1 as x tends to 0, from our above sin(x) <x, and sinx >=0 for x>=0 giving sinx tends to zero. Thus cos^2x tends to 1, and so does cos x(as it is certainly positive). Now take e>0 and x such that cosx>1e. We want to show (1e)x<sinx.
Proof: consider a line inclined at angle x, and concentric circles at radius 1e and 1. Then (1e)x is the arc length between the x axis and the line. Now this distance is bounded above by (1e)(1cosxe) +(1e)sinx=(1e)(sinx+(1e)cosx)<sinx + (1ecosx)<sinx. This is from drawing a rectangle on the smaller circle bounding the arc, and then the bound is clear, as if not a circle would not be the shape of maximal area by puting this rectangle into a circle.
Edit: I messed this up again  the bound should be (1e)(sinx+1cosx). Back to the drawing board...geometry clearly isnt my forte

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Re: Wanted: Elegant proof of sin(x)/x limit
Qaanol wrote:I know of a simple way to show that (1cos(x))/x → 0, but I have not yet found a correspondingly transparent illustration for sin(x)/x.
Have you posted this proof already? If not, then please do? I am bad at spotting posts.
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Re: Wanted: Elegant proof of sin(x)/x limit
Consider the angle between the chord and the upright.
(Hint: bisect the sector.)
(Hint: bisect the sector.)
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