Re: sin x / x tends to 1...
- From: "Proginoskes" <CCHeckman@xxxxxxxxx>
- Date: 7 Sep 2005 15:38:16 -0700
bill wrote:
> 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. 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,
> > --
> > Dr Darren Wilkinson
> > mailto:d.j.wilkinson@xxxxxxxxxxxxxxx
> > http://www.staff.ncl.ac.uk/d.j.wilkinson/
>
>
> Sin(x)/x = 1 - x^2/3! + ...
> Cos(x) = 1 - x^2/2! + ...
>
> As long as x > 0, cosx =/= sinx. So is it correct to say that sinx/x
> approaches cosx as x appropaches zero? I say no!
Well, Wilkinson isn't saying that cos(x) = sin(x)/x. What's actually
true is that
cos(x) < sin(x)/x < 1,
and as x approaches 0, cos(x) approaches 1. The sandwich/squeeze
theorem says that under these circumstances, lim (sin(x)/x) = 1.
It's possible for two different functions to have the same limit as x
approaches 0; for instance 3x and x^2. This does NOT make 3x = x^2.
--- Christopher Heckman
.
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