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

On Sat, 03 Sep 2005 22:53:37 GMT, "Liviu" <lab2k1@xxxxxxxxxxx> wrote:

>
>"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> wrote in message
>news:554kh1h96cv6uak5i2i3nhcebdlo6l0ng8@xxxxxxxxxx
>> On Sat, 3 Sep 2005 17:32:57 +0000 (UTC), Jim Spriggs
>> <jim.sprigs@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> 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.
>>>
>>>_If_ you are willing to accept certain properties of area, for example
>>>that the area of the sector of a circle exceeds the area of the triangle
>>>inscribed in it, then proceed as follows:
>>>
>>>If 0 < x < pi/2 then
>>>
>>> sin x < x < tan x
>>
>> The problem with that is that [.. snip ..]
>>
>> Or if we just look at arc lengths instead of areas
>> I see how to show that sin(x) < x, but I don't see
>> how to show that the _length_ of an arc is less than
>> _anything_ (unless we're willing to believe that it's
>> the limit of the lengths of inscribed polygons or
>> something...) [.. snip ..]
>
>If one accepts, or considers "easy" to prove in this context, that (a) arcs
>are convex and (b) the perimeter of a convex closed curve is less than that
>of any other one enclosing it,

Yes, it certainly follows if you accept that. That doesn't strike me
as nearly as obviously "acceptable" as the fact that a straight line
is the shortest distance between two points.

>then I think x < tan(x) could follow. Rough
>sketch, with no pretense to rigor... Let O be the center of the unit circle,
>A and X points on the circle such as the angle XOA is x, Y the projection of
>X onto OA, Z the intercept of the tangent at X with OA. Then XY = sin(x), XZ
>= tan(x) and the arclength (XA) = x is between them [ using assumption (b)
>above, after completing closed curves by reflecting X across OA and A, Z
>across XY ] i.e. sin(x) < x < tan(x).
>
>Liviu
>
>
>
>


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

David C. Ullrich
.

Relevant Pages

  • Re: sin x / x tends to 1...
    ... > how to show that the _length_ of an arc is less than ... are convex and the perimeter of a convex closed curve is less than that ... Let O be the center of the unit circle, ...
    (sci.math)
  • Re: sin x / x tends to 1...
    ... >>>that the area of the sector of a circle exceeds the area of the triangle ... I think areas are simpler to deal with than arc lengths. ... >> David C. Ullrich ...
    (sci.math)
  • Re: hi
    ... > I look for a line segment from the center to the circle. ... >> that the arc length I speak of is the smaller of the two arc lengths ... >> if there is no way to avoid doing this by some approximation method ... as of yet i am any idiot becuse that's the only way i can ...
    (sci.math)