Limit of sinx divided by x, as x approaches 0

lim [sin(x)]/x
x→0

the answer is 1, as I can see on my graphing calculator, but why?

L'Hopital's rule
lim x-->0 =
cos x/1 = cos 0 /1 = 1

Draw a unit circle on xOy.
r=OA=OB=1 (O(0,0))
∠AOB=x (x＞0 and very small)
draw AH⊥OB at H.
=> AH=sinx; OH=cosx, ⌒AB=1*x=x (⌒for arc).
passing B draw OB's perpendicular line points OA
at C => BC=tanx => BC=tanx => △AOB＜OAB＜△OBC
(OAB is a sector).
=> 1/2*1*1*sinx＜pi*1²*x/2pi＜1/2*1*tanx
=> sinx＜x＜tanx=sinx/cosx => 1＜x/sinx＜1/cosx
=> 1＞sinx/x＞cosx, i.e cosx＜sinx/x＜1
=> when x tends 0, cosx->1, 1->1, sinx/x->1
x tends to 0 from right.
when x tends 0 from left:
x＜0; sinx＜0, cosx＞0
similarly |sinx|＜|x|＜|tanx|=|sinx/cosx|
=> 1＜|x/sinx|＜1/|cosx|
=> 1＜x/sinx＜cosx
∴;lim x->0 (sinx/x)=1
My english is not well...here is a picture
http://p13.freep.cn/p.aspx?u=v20_p13_pho…

we have
sin(x) =x- x^/3 + x^5 - x^7 + x^9 - ...................
[sin(x)]/x =1-x^2 + x^4 -x^6 -x^8 -.......................

lim [sin(x)]/x = lim [1-x^2 + x^4 -x^6 -x^8 -.......................]
x→0 x→0

lim [sin(x)]/x =1
x→0

Note : only 'wen tian qian' is valid , you cannot use L'H or a series as they both

ASSUME the limit is known to be 1..ie. that the derivative is cos x !!