求极限方法总结全

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≠0.2、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,a.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(a.b>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+1lny+1y=1例题6、limx→π21-sinxα+β1-sinxα1-sinxβα.β>0解：令y=1-sinxlimx→π21-sinxα+β1-sinxα1-sinxβ=limy→0+1-1-yα+β1-1-yα1-1-yβ=limy→0+yα+βαyβy=α+βαβ（4）1∞型求极限例题1、limx→π4tanxtan2x解：解法一（等价无穷小）：limx→π4tanxtan2x=elimx→π4tan2xlntanx=elimx→π4tan2xln1+tanx-1=elimx→π4tan2xtanx-1=elimx→π42tanx1-tanx2tanx-1=elimx→π4-2tanx1+tanx=e-1解法二（重要极限）：limx→π4tanxtan2x=limx→π41+tanx-11tanx-1tan2xtanx-1=elimx→π4tan2xtanx-1=elimx→π42tanx1-tanx2tanx-1=elimx→π4-2tanx1+tanx=e-1（5）夹逼定理（主要适用于数列）例题1、limn→∞1n+2n+3n+4n1n解：4n≤1n+2n+3n+4n≤4×4n所以limn→∞1n+2n+3n+4n1n=4推广：ai>0i=1,2,3……mlimn→∞a1n+a2n+a3n+…+amn1n=max1≤i≤mai例题2、limx→0x1x解：1x-1≤1x≤1xx>01-x≤x1x≤1所以x→0+limx→0x1x=1x<01-x≥x1x≥1所以x→0-limx→0x1x=1例题3、limn→∞32×55×78×?×2n+13n-1解：2n+13n-1≤2n+13nn≥20≤32×55×78×?×2n+13n-1≤32×66×89×?×2n+13n=n+1223n-2limn→∞n+1223n-2=0所以limn→∞32×55×78×?×2n+13n-1=0例题4、limn→∞k=n2n+121klimn→∞k=n2n+121k=limn→∞1n2+1n2+1+?+1n+122n+2n+12≤xn≤2n+2n2所以limn→∞xn=2例题5、limn→∞k=1nnk+1-1k解：nk≤nk+1≤n+1kn≤nk+11k≤n+11n+1≤nk+1-1k≤1n所以nn+1≤k=1nnk+1-1k≤nnlimn→∞k=1nnk+1-1k=1（6）单调有界定理例题1、limn→∞32×55×78×?×2n+13n-1解：xn=xn-1×2n+13n-1≤xn-1???*xn单调递减0≤xn极限存在，记为A由（*）n→∞求极限得：A=23A所以A=0例题2、x0=1xn+1=2xn求limn→∞xn解：xn+1-xn=2xn-2xn-1=2xn-xn-12xn+2xn-1x1-x0=2-1>0xn单调递增xn+1=2xn<2xn+1所以xn+12-2xn+1<000xn+1=a1+xna+xn(a>1)求极限limn→∞xn解：xn+1-xn=a1+xna+xn-a1+xn-1a+xn-1=a2-axn-xn-1a+xna+xn-1x2-x1=a-x12a+x1当x1>ax2-x1<0xn↓当01xn+1=11+xn求极限limn→∞xn解：xn+1-xn=xn-1-xn1+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-x1<0所以x2n+1单调递减，同理，x2n单调递增有因为00)解：limx→+∞xneλx=limx→+∞nxn-1λeλx=limx→+∞nn-1xn-2λ2eλx=?=limx→+∞n!λneλx=0例题4、求limx→0+xnlnxn>0解：limx→0+xnlnx=limx→0+lnxx-n=limx→0+1x-nx-n-1=limx→0+-xnn=0（9）利用函数的图像通过对求解极限方法的研究，我们对极限有了进一步的了解。极限方法是研究变量的一种基本方法，在以后的学习过程中，极限仍然起着重要的作用，因此学习、掌握极限是十分必要的。相信通过对极限的学习总结，我们在今后的学习中能更进一步。