求极限lim(x趋于无穷）cos√(x+1)-cos√x，用夹逼准则 怎么用夹逼准则证明cos（1/n）在n趋向于无穷的极限
\u6c42\u6781\u9650lim(x\u8d8b\u4e8e\u65e0\u7a77\uff09cos\u221a(x+1)-cos\u221ax\uff0c\u7528\u5939\u903c\u51c6\u5219cos\u221a(x+1) - cos\u221ax = - 2 sin [ (\u221a(x+1)-\u221ax)/2] sin[ (\u221a(x+1)+\u221ax)/2]
- 2 | sin[ (\u221a(x+1)+\u221ax)/2] | \u2264 cos\u221a(x+1) - cos\u221ax \u2264 2 | sin[ (\u221a(x+1)+\u221ax)/2] |
\u5f53 x->+\u221e\u65f6 \u221a(x+1)-\u221ax = 1/ (\u221a(x+1)+\u221ax ) -> 0, sin [ (\u221a(x+1)-\u221ax)/2] -> 0
\u4e8e\u662f - 2 | sin[ (\u221a(x+1)+\u221ax)/2] | ->0 \u4e14 2 | sin[ (\u221a(x+1)+\u221ax)/2] | ->0
\u7531\u8feb\u655b\u51c6\u5219(\u5939\u903c\u51c6\u5219),
lim (x->+\u221e ) [ cos\u221a(x+1) - cos\u221ax ] = 0
\u56e0\u4e3a\u5f53x>0\u65f61>cosx>1\uff0dx^2/2
\u5173\u4e8ecosx>1\uff0dx^2/2\u7684\u8bc1\u660e\uff0c\u53ef\u4ee5\u5229\u7528\u5bfc\u6570\u7684\u65b9\u6cd5\u8bc1\u660e
\u5f53x\u8d8b\u4e8e0\u662f\uff0ccosx\u8d8b\u4e8e1.\u7b49\u4ef7\u4e8en\u8d8b\u4e8e\u221e\u65f6\uff0ccos(1/n)\u8d8b\u4e8e1
\u53e6\u6cd5\uff1a
\u5f530<x<\u03c0/2\u65f6
sinx<x<tanx
\u6240\u4ee5sinx/x<1<tanx/x
\u4e24\u8fb9\u540c\u4e58\u4ee5cosx\u5f97\u5230
sin(2x)/(2x)<cosx<sinx/x
\u5f53x\u8d8b\u4e8e0\u65f6,sinx/x\u4e0esin(2x)/(2x)\u90fd\u8d8b\u4e8e1,\u6545cos(x)\u8d8b\u4e8e1
cosA-cosbB把其中的A化成 [(A+B)+(A-B)]/2，B化成 [(A+B)-(A-B)]/2这种形式！
所以按照这个思路做下去！
0<|cos√(x+1)-cos√x|=........<=0 故有迫敛性可知极限是0，建议以后这个定理称之为 迫敛性吧，你说的那个名字实在是难听，呵呵！
设t=(√(x+1)+√x) (√(x+1)-√x)/2=1/(2(√(x+1)+√x))=1/(2t)
cos√(x+1)-cos√x
=-2sin((√(x+1)+√x)/2)sin((√(x+1)-√x)/2)
=-2sin(t/2)*sin(1/(2t))
<=|2sin(t/2)*sin(1/(2t))|
=2|sin(t/2)||sin(1/(2t))|
<=2|sin(1/(2t))|
即-2sin(1/(2t))<=cos√(x+1)-cos√x<=2sin(1/(2t))
当x趋于无穷时，1/2t→0，sin(1/2t)→0，0<=lim(x趋于无穷）cos√(x+1)-cos√x<=0
所以lim(x趋于无穷）cos√(x+1)-cos√x=0
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