文档介绍：极限求解总结1、极限运算法则设limn→∞an=a,limn→∞bn=b,则limn→∞(an±bn)=limn→∞an±limn→∞bn=a±b;limn→∞anbn=limn→∞anlimn→∞bn=ab;limn→∞anbn=limn→∞anlimn→∞bn=abb≠、函数极限与数列极限的关系如果极限limx→x0f(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:xn≠x0n∈N+,那么相应的函数值数列f(x)必收敛,且limn→∞f(xn)=limx→x0f(x)3、定理有限个无穷小的和也是无穷小;有界函数与无穷小的乘积是无穷小;4、推论常数与无穷小的乘积是无穷小;有限个无穷小的乘积也是无穷小;如果limf(x)存在,而c为常数,则limcf(x)=climf(x)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limx→x0gx=u0,limu→u0f(u)=A,且存在δ0>0,当x∈Ux0,δ0时,有g(x)≠u0,则limx→x0f[gx]=limu→u0f(u)=A6、夹逼准则如果当x∈Ux0,r(或x>M)时,g(x)≤f(x)≤h(x)limx→x0(x→∞)g(x)=A,limx→x0(x→∞)h(x)=A那么limx→x0(x→∞)f(x)存在,且等于A7、两个重要极限limx→0sinxx=1limx→∞1+1xx=e8、求解极限的方法(1)提取因式法例题1、求极限limx→0ex+e-x-2x2解:limx→0ex+e-x-2x2=limx→0e-x(e2x-2ex+1)x2=limx→0e-xex-1x2=1例题2、求极限limx→0ax2-bx2ax-bx2a≠b,>0解:limx→0ax2-bx2ax-bx2=limx→0bx2abx2-1b2xabx-12=limx→0bx2-2xx2lnabxlnab2=1lnab例题3、求极限limx→+∞xpa1x-a1x+1a>0,a≠1解: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、limx→πsinmxsinnx解:令x=y+πlimx→πsinmxsinnx=limy→0sinmy+mπsinny+nπ=-1m-nlimy→0sinmysinny=-1m-nmn例题2、limx→1x1m-1x1n-1解:令x=y+1limx→1x1m-1x1n-1=limx→1(1+y)1m-1(1+y)1n-1=nm例题3、limx→+∞x2x2+x-3x3+x2解:令y=1xlimx→+∞x2x2+x-3x3+x2=limy→0+1y2+1y-31y3+1y2=limy→0+1+y-31+yy=16(3)等价无穷小替换法x→0sinx~x~sin-1xtanx~x~tan-1xex-1~x~ln1+xax-1~xlna1-cosx~x221+xα-1~αx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、limx→0ax+bx21x(>0)解:limx→0ax+bx21x=elimx→01xlnax+bx2=elimx→01xln1+ax+bx-22=elimx→0ax-1+bx-12x=ab例题2、limx→+∞ln1+eaxln1+bx(a>0)解:limx→+∞ln1+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、limx→0lnsinx2+ex-xlnx2+e2x-2x解:limx→0lnsinx2+ex-xlnx2+e2x-2x=limx→0lnsinx2+ex-xlnx2+e2x-2x=limx→0lnsinx2ex+1lnx2e2x+1=limx→0sinx2e2xx2ex=1例题4、limx→0ex-esinxx-sinx解:limx→0ex-esinxx-sinx=limx→0esinxex-sinx-1x-sinx=limx→0esinxx-sinxx-sinx=1例题5、limx→1xx-1x-1解:limx→1xx-1x-1=limx→1exlnx-1x-1=limx→1xlnxx-1令y=x-1原式=limy→0y+