文档介绍：(两边夹定理)定理Ⅰ如果数列nnyx,及nz满足下列条件:,lim,lim)2()3,2,1()1(azaynzxynnnnnnn??????????那末数列nx的极限存在,且axnn???,,azaynn??因为使得,0,0,021?????NN?一极限存在准则,1????ayNnn时恒有当},,max{21NNN?取,??????ayan即,2????azNnn时恒有当,??????azan上两式同时成立,恒有时当,Nn?,????????azxyannn,成立即???????上述数列极限存在的准则可以推广到函数的极限准则Ⅰ′如果当)(00xUx??(或Mx?)时,有,)(lim,)(lim)2(),()()()1()()(00AxhAxgxhxfxgxxxxxx??????????那末)(lim)(0xfxxx???存在,且等于A .注意:.,'I例1).12111(lim222nnnnn?????????求解,11112222????????nnnnnnnn??nnnnnn111limlim2???????又,1?22111lim1limnnnnn???????,1?)12111(lim222?????????nnnnn?x1x2x3x1?,121???????nnxxxx单调增加,121???????nnxxxx单调减少单调数列准则Ⅱ:AM证,1nnxx??显然??;是单调递增的nx?,331??x?又,3?kx假定kkxx???3133??,3???;是有界的nx?.lim存在nnx???,31nnxx????,321nnxx???),3(limlim21nnnnxx???????,32AA??2131,2131????AA解得(舍去).2131lim?????nnx.)n例2(的极限存在式重根证明数列333?????nxAC二、两个重要极限1、1sinlim0??xxxxoBD)20(,,?????xxAOBO圆心角设单位圆,tan,,sinACxABxBDx????,得作单位圆的切线,xOAB的圆心角为扇形,BDOAB的高为?,tansinxxx???,1sincos??????x,20时当???xxxcos11cos0????2sin22x?2)2(2x?,22x?,02lim20??xx?,0)cos1(lim0????xx,1coslim0???xx,11lim0??x?又10???xxxsinlim246810--10-8-6-4-2-:利用变量代换可导出上;1)()(sinlim0)(??xxx???