Limit of sinx divided by x, as x approaches 0
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Limit of sinx divided by x, as x approaches 0

[From: ] [author: ] [Date: 12-09-04] [Hit: ]
freep.cn/p.aspx?u=v20_p13_pho…-we have sin(x) =x- x^/3 + x^5 - x^7 + x^9 -.........
lim [sin(x)]/x
x→0

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

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

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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…

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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

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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 !!
1
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