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1、极限求解总结1、极限运算法则设limnan=a,limnbn=b,则(1) limn(anbn)=limnanlimnbn=ab;(2) limnanbn=limnanlimnbn=ab;(3) limnanbn=limnanlimnbn=abb0.2、函数极限与数列极限的关系如果极限limxx0f(x)存在,xn为函数f(x)的定义域内任一收敛于x0的数列,且满足:xnx0nN+,那么相应的函数值数列f(x)必收敛,且limnf(xn)=limxx0f(x)3、定理(1) 有限个无穷小的和也是无穷小;(2) 有界函数与无穷小的乘积是无穷小;4、推论(1) 常数与无穷小的乘积是无穷小;(2)
2、有限个无穷小的乘积也是无穷小;(3) 如果limf(x)存在,而c为常数,则 limcf(x)=climf(x)(4) 如果limf(x)存在,而n是正整数,则 limf(x)n=limf(x)n5、复合函数的极限运算法则设函数y=fgx是由函数u=gx与函数y=f(u)复合而成的,y=fgx在点x0的某去心领域内有定义,若limxx0gx=u0,limuu0f(u)=A,且存在00,当xUx0,0时,有g(x)u0,则limxx0fgx=limuu0f(u)=A6、夹逼准则 如果(1) 当xUx0,r(或xM)时,g(x)f(x)h(x)(2) limxx0(x)g(x)=A, limxx0
3、(x)h(x)=A那么limxx0(x)f(x)存在,且等于A7、两个重要极限(1) limx0sinxx=1(2) limx1+1xx=e8、求解极限的方法(1)提取因式法例题1、求极限limx0ex+e-x-2x2解:limx0ex+e-x-2x2=limx0e-x(e2x-2ex+1)x2=limx0e-xex-1x2=1例题2、求极限limx0ax2-bx2ax-bx2ab,a.b0解:limx0ax2-bx2ax-bx2=limx0bx2abx2-1b2xabx-12=limx0bx2-2xx2lnabxlnab2=1lnab例题3、求极限limx+xpa1x-a1x+1a0,a1解
4、:limx+xpa1x+1a1x(x+1)-1=limx+xpa1x+11x(x+1)lna=limx+xpx(x+1)a1x+1lna=limx+xp-21+1xa1x+1lna=(2)变量替换法(将不一般的变化趋势转化为普通的变化趋势)例题1、limxsinmxsinnx解:令x=y+limxsinmxsinnx=limy0sinmy+msinny+n=-1m-nlimy0sinmysinny=-1m-nmn例题2、limx1x1m-1x1n-1解:令x=y+1limx1x1m-1x1n-1=limx1(1+y)1m-1(1+y)1n-1=nm例题3、limx+x2x2+x-3x3+x2解
5、:令y=1xlimx+x2x2+x-3x3+x2=limy0+1y2+1y-31y3+1y2=limy0+1+y-31+yy=16(3)等价无穷小替换法x0 sinxxsin-1x tanxxtan-1x ex-1xln1+x ax-1xlna 1-cosxx22 1+x-1x注:若原函数与x互为等价无穷小,则反函数也与x互为等价无穷小例题1、limx0ax+bx21x(a.b0)解:limx0ax+bx21x=elimx01xlnax+bx2=elimx01xln1+ax+bx-22=elimx0ax-1+bx-12x=ab例题2、limx+ln1+eaxln1+bx(a0)解:limx+l
6、n1+eaxln1+bx=limx+ln1+eaxbx=limx+bxlneaxe-ax+1=limx+bxlneax+lne-ax+1=limx+bxax+lne-ax+1=ab+limx+blne-ax+1x=ab例题3、limx0lnsinx2+ex-xlnx2+e2x-2x解:limx0lnsinx2+ex-xlnx2+e2x-2x=limx0lnsinx2+ex-xlnx2+e2x-2x=limx0lnsinx2ex+1lnx2e2x+1=limx0sinx2e2xx2ex=1例题4、limx0ex-esinxx-sinx解:limx0ex-esinxx-sinx=limx0esin
7、xex-sinx-1x-sinx=limx0esinxx-sinxx-sinx=1例题5、limx1xx-1x-1解:limx1xx-1x-1=limx1exlnx-1x-1=limx1xlnxx-1令y=x-1原式=limy0y+1lny+1y=1例题6、limx21-sinx+1-sinx1-sinx.0解:令y=1-sinxlimx21-sinx+1-sinx1-sinx=limy0+1-1-y+1-1-y1-1-y=limy0+y+yy=+(4)1型求极限例题1、limx4tanxtan2x解:解法一(等价无穷小):limx4tanxtan2x=elimx4tan2xlntanx=el
8、imx4tan2xln1+tanx-1=elimx4tan2xtanx-1=elimx42tanx1-tanx2tanx-1=elimx4-2tanx1+tanx=e-1解法二(重要极限):limx4tanxtan2x=limx41+tanx-11tanx-1tan2xtanx-1=elimx4tan2xtanx-1=elimx42tanx1-tanx2tanx-1=elimx4-2tanx1+tanx=e-1(5)夹逼定理(主要适用于数列)例题1、limn1n+2n+3n+4n1n解:4n1n+2n+3n+4n44n所以limn1n+2n+3n+4n1n=4推广:ai0 i=1,2,3mli
9、mna1n+a2n+a3n+amn1n=max1imai例题2、limx0x1x解:1x-11x1x1) x0 1-xx1x1所以x0+ limx0x1x=12) x0xn单调递增xn+1=2xn2xn+1所以xn+12-2xn+100xn+10 xn+1=a1+xna+xn(a1)求极限limnxn解:xn+1-xn=a1+xna+xn-a1+xn-1a+xn-1=a2-axn-xn-1a+xna+xn-1x2-x1=a-x12a+x1当x1a x2-x10 xn 当0x1a xn所以0xn+1=a1+xna+xn1 xn+1=11+xn 求极限limnxn解:xn+1-xn=xn-1-xn
10、1+xn1+xn-1 (整体无单调性)x2n+1-x2n-1=11+x2n-11+x2n-2=x2n-2-x2n1+x2n1+x2n-2=x2n-1-x2n-31+x2n1+x2n-21+x2n-11+x2n-3x3-x1=11+x2-x10所以x2n+1单调递减,同理,x2n单调递增有因为0xn1(n2)故limnx2n+1和limnx2n均存在,分别记为A,Bx2n+1=11+x2n x2n=11+x2n-1 即A=11+B B=11+A解得 A=B=5-12所以 limnxn=5-12(7)泰勒公式法例题1、设f有n阶连续导数n2fkx0=0 k=1,2,n-1fnx00 nRfx0+h-fx0=hfx0+h 0=h1证明:limh0h=n11-n证明:fx0+h=fx0+fx0h+f3x02!h2+fn-1x0n-2!hn-2+fnn-1!hn-1即fx0+h=fnn-1!hn-1 x0x0+hfx0+h= fx0+fnn!hnfx0+h-fx0=hnfnn! x00)解:limx+xnex=limx+nxn-1ex=l