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

On Sat, 3 Sep 2005 13:22:46 +0000 (UTC), Darren J Wilkinson
<d.j.wilkinson@xxxxxxxxx> 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.

Not really. Why not? See a few paragraphs down, after some
disclaimers:

First I should say that it seems to me that the simplest
thing really is to define sin(x) by the power series. Then
the proof that sin(x)/x -> 1 is very simple (it's not clear
to me whether you said in a post below that this was not
clear to you), and it's not too hard to show that this
turns out to be the same as a geometric definition.

But if you say that's not elementary enough fine. Let's
talk about sin defined "geometrically", and let's ignore
the fact that just _defining_ what the area of a region
_is_ takes some not-so-elementary work; let's assume,
as in a typical elementary context, that we know what
areas are.

(Um, when I say let's assume we know what area is I
don't mean we're going to assume we know that the
area of a circle of radius 1 is pi, I just mean that
we're going to assume that we know what the _notion_
of "area" means.)

Ok. We do have to assume that the area of a circular
sector is proportional to the angle. To prove that we'd
first have to define "area", and that would take us
out of absolutely elementary things. It's certainly
intuitively plausible from our intuitive notion of
"area".

We assume that the area of a circular sector is
proportional to the angle. We define the sine of
an _angle_ geometrically. We define a radian to
be that angle such that a circular sector with
radius 1 and opening one radian has area 1.
If we're justifying things then the existence
of such an angle depends on continuity and the
intermediate value theorem. But we're not
worrying about that, lest things become
non-elementary - the point is we define one
radian in terms of a certain _area_, instead
of in terms of a certain arc length. If you're
worried about how to show that that's the same
as a radian defined in terms of arc length
don't worry about that - say it's "aradian"
defined in terms of area and "radian" defined
in terms of arc length - we won't need the
fact that 1 aradian = 1 radian.

Now we define the sine of an _angle_ geometrically,
and then if x is a _number_ we say that sin(x) is
the sine of x radians. Now draw that picture you
see in calculus books.

That picture contains a little triangle contained
in a circular sector, which is contained in a
larger triangle. If you consider the areas of those
three sets you see

sin(x) cos(x) <= x <= tan(x)

(the sector has area x by our definition of
"radian"!) Hence

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

and we're done.

There's a lot there that was not "rigorous" - you
said you wanted elementary. But the sort of, um,
circularity you're concerned with here simply
doesn't come up - we didn't need to know the
area of a circle.

(We could show the area of a circle was pi r^2
by defining pi to be the number of radians in
180 degrees. Now to show that the circumference
is what it is we finally need to show that
"aradians" are the same as radians...)

>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. 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?
>
>Regards,


************************

David C. Ullrich
.

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