Re: sin x / x tends to 1...


Darren J Wilkinson wrote:
> I've a question about the limit of sin x / x as x tends to zero. Of
> course, it's 1 (I think), but I've never seen a satisfactory proof. The
> proof I was given, and the proofs I can find in standard texts all rely
> on knowing the area of a circular sector. However, to know the area of a
> circular sector, one must know the area of a circle. All the derivations
> I know for the area of a circle make use (either directly or indirectly)
> on the sin x / x limit, and there lies my disatisfaction.

Didn't Euclid prove that the area of a circle is pi r^2, where r is
the
radius, by purely geometric means? Once you have that, then the area of
a sector follows immediately; you multiply the total area by theta/(2
pi).

> Of course it's
> easy to get the upper bound of one, and I'm happy to use the area
> argument to establish the existance of a limit. However, it seems to be
> surprisingly awkward to establish the obvious lower bounds (such as cos
> x) using elementary arguments. Does anyone know a nice proof?

I didn't read the entire thread; it's become too big for a first-time
person to get through it all in a reasonable amount of time, so please
forgive me if this proof was posted by someone else.

Stewart's book uses the following, which only uses lengths of line
segments and arcs, when theta is strictly between 0 and pi/2:

(1) Construct a triangle OAD, where O is the origin, A is on the
x-axis, the angle DOA is theta, the angle OAD is pi/2 radians, and the
distance OA is 1.

(2) Draw the circle centered at O with radius 1. It will cross the line
segment OD at point B.

(3) Drop a perpendicular from B down to the x-axis, to get to point C.

(4) Draw the line tangent to the circle at B, and let the point where
it crosses line segment DA be E.

(5) The length of the arc AB is theta, |BC| = sin(theta). (|*| means
"leangth of *".) From the diagram, we get

|BC| < |AB| (AB is the hypotenuse of triangle ABC)
< |arc AB|. This implies that
sin (theta) < theta, so sin(theta)/theta < 1.

(6) |arc AB| < |BE| + |EA| ("obvious")
< |AE| + |ED| (ED is the hypotenuse of triangle BED)
= |AD| = tan(theta).

Since the length of AB is theta, this proves that theta < tan(theta),
or

theta < sin(theta)/cos(theta), so

cos(theta) < sin(theta)/theta.

And now we use the squeeze/sandwich theorem and let theta approach 0,
using the combined fact

cos(theta) < sin(theta)/theta < 1.

--- Christopher Heckman

.

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