求极限方法总结全

极限求解总结1、极限运算法则设lim n →∞a a =a ,lim n →∞a a =a ,则(1) lim n →∞(a a ±a a )=lim n →∞a a ±lim n →∞a a =a ±a ;(2) lim n →∞a a a a =lim n →∞a a lim n →∞a a =aa ;(3) limn →∞a aa a=lim n →∞a a lim n →∞a a=a a(a ≠0).2、函数极限与数列极限的关系如果极限lim x →a 0a (a )存在，{a a }为函数a (a )的定义域内任一收敛于a 0的数列，且满足:a a ≠a 0(a ∈a +)，那么相应的函数值数列{a (a )}必收敛，且lim a →∞a (a a )=lim a →a 0a (a )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） 有限个无穷小的乘积也是无穷小；（3）如果lim a(a)存在，而c为常数，则lim[aa(a)]=a lim a(a)（4）如果lim a(a)存在，而n是正整数，则lim[a(a)]a=[lim a(a)]a5、复合函数的极限运算法则设函数y=a[a(a)]是由函数u=a(a)与函数y=a(a)复合而成的，y=a[a(a)]在点a0的某去心领域内有定义，若lima→a0a(a)=a0，lima→a0a(a)=a，且存在a0>0，当x∈U(a0,a0)时，有a(a)≠a0，则lima→a0a[a(a)]=lima→a0a(a)=a6、夹逼准则如果(1)当x∈U(a0,a)(或|a|>M)时,g(x)≤a(a)≤h(x)(2)lima→a0(a→∞)a(a)=a,lima→a0(a→∞)a(a)=a那么lima→a0(a→∞)a(a)存在，且等于A 7、两个重要极限（1）lima→0sin aa=1（2）limx→∞(1+1x)x=a8、求解极限的方法（1）提取因式法例题1、求极限lima→0a a+a−a−2a2解：lima→0a a+a−a−2a2=lima→0a−a(a2a−2a a+1)a2=lima→0a−a(a a−1a)2=1例题2、求极限lima→0a a2−a a2(a a−a a)2(a≠a,a.a>0)解：lima→0a a2−a a2(a a−a a)2=lima→0a a2[(aa)a2−1]a2a[(aa)a−1]2=lima→0a a2−2aa2ln aa(a ln aa)2=1lnaa例题3、求极限lima→+∞a a(a1a−a1a+1)(a>0,a≠1)解：lima→+∞a a a1a+1(a1a(a+1)−1)=lima→+∞a a a1a+11a(a+1)ln a=lim a→+∞a aa(a+1)a1a+1ln a=lima→+∞a a−21+1aa1a+1ln a=（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、lima→a sin(aa) sin(aa)解：令x=y+πlim a→a sin(aa)sin(aa)=lima→0sin(aa+aa)sin(aa+aa)=(−1)a−a lima→0sin aasin aa=(−1)a−aaa例题2、lima→1a1a−1 a1a−1解：令x=y+1lim a→1a1a−1a1a−1=lima→1(1+a)1a−1(1+a)1a−1=aa例题3、lima→+∞a2√−√3解：令y=1alim a→+∞a2√+−√+3=lima→0+√1a2+1a−√1a3+1a23=lima→0+√1+a−√1+a3a=16（3）等价无穷小替换法x→0sin a~a~sin−1a tan a~a~tan−1aa a−1~a~ln(1+a)a a−1~a ln a1−cos a~a22(1+a)a−1~aa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、lima→0(a a+a a2)1a(a.a>0)解：lima→0(a a+a a2)1a=a lim a→01a ln aa+a a2=a lim a→01aln(1+aa+a a−22)=a lim a→0(a a−1)+(a a−1)2a=√aa例题2、lima→+∞ln(1+a aa)ln(1+ba)(a>0)解：lima→+∞ln(1+a aa)ln(1+ba)=lima→+∞ln(1+a aa)aa=lim a→+∞aaln[a aa(a−aa+1)]=lima→+∞aa[ln a aa+ln(a−aa+1)]=lim a→+∞aa[aa+ln(a−aa+1)]=aa+lima→+∞a ln(a−aa+1)a=aa例题3、lima→0ln((sin a)2+a a)−a ln(a2+a2a)−2a解：lima→0ln((sin a)2+a a)−aln(a2+a2a)−2a=lima→0ln((sin a)2+a a)−aln(a2+a2a)−2a=lima→0ln((sin a)2a a+1)ln(a2a2a+1)=lima→0(sin a)2a2aa2a a=1例题4、lima→0a a−a sin a a−sin a解：lima→0a a−a sin aa−sin a=lima→0a sin a(a a−sin a−1)a−sin a=lima→0a sin a(a−sin a)a−sin a=1例题5、lima→1a a−1 a−1解：lima→1a a−1a−1=lima→1a a ln a−1a−1=lima→1a ln aa−1令y=x-1原式=lima→0(a+1)ln(a+1)a=1例题6、lima→21−(sin a)a+a(())(())(a.a>0)解：令y=1−sin alim a →a21−(sin a )a +a√(())(())=lim a →0+1−(1−a )a +a√[()][()]=lim a →0+a (a +a )√aaaa=a +a√aa（a）a ∞型求极限例题1、lim a →a 4(tan a )tan 2a 解：解法一（等价无穷小）：lim a →a 4(tan a )tan 2a=alim a →a 4(tan 2a )ln (tan a )=alim a →a 4(tan 2a )ln [1+(tan a −1)]=a lim a →a 4(tan 2a )(tan a −1)=alim a →a42tan a1−(tan a )2(tan a −1)=alim a →a4−2tan a 1+tan a =a −1解法二（重要极限）：lim a →a 4(tan a )tan 2a=lim a →a 4[1+(tan a −1)]1tan a −1tan 2a (tan a −1)=alim a →a 4(tan 2a )(tan a −1)=alim a →a42tan a1−(tan a )2(tan a −1)=alim a →a 4−2tan a 1+tan a =a −1（5）夹逼定理（主要适用于数列） 例题1、lim a →∞(1a+2a+3a+4a )1a解：4a ≤1a +2a +3a +4a ≤4×4a 所以lim a →∞(1a+2a+3a+4a )1a=4推广：a a>0a=1,2,3……mlim a→∞(a1a+a2a+a3a+⋯+aaa)1a=max1≤a≤a{aa}例题2、lima→0a[1a]解：1a −1≤[1a]≤1a1)x>01−x≤x[1a]≤1所以x→0+lima→0a[1a]=12)x<01−x≥x[1a]≥1所以x→0−lima→0a[1a]=1例题3、lima→∞32×55×78×?×2a+13a−1解：2a+13a−1≤2(a+1)3a(a≥2)0≤3×5×7×?×2a+1−≤3×6×8×?×2(a+1)=a+1(2)a−2lima→∞a+12(23)a−2=0所以lima→∞32×55×78×?×2a+13a−1=0例题4、lima→∞∑1√(a+1)2a=a2lim a→∞∑1√a(a+1)2a=a2=lima→∞[1√+1√a1+?1√()]2a+2√()≤a a≤2a+2√所以lima→∞a a=2例题5、lima→∞∑(a a+1)−1a aa=1解：a a≤a a+1≤(a+1)a a≤(a a+1)1a≤a+11 a+1≤(a a+1)−1a≤1a所以aa+1≤∑(a a+1)−1aaa=1≤aalima→∞∑(a a+1)−1aaa=1=1（6）单调有界定理例题1、lima→∞32×55×78×?×2a+13a−1解：a a=a a−1×2a+13a−1≤a a−1???(∗){aa }单调递减0≤aa极限存在，记为A由（*）a→∞求极限得：A=23A所以A=0例题2、a0=1a a+1=√2a a求lima→∞a a解：a a+1−a a=√2a a−√2a a−1=(a a−1a a−1a1−a0=√2−1>0{aa}单调递增a a+1=√2a a<√2a a+1所以(a a+1)2−2a a+1<00<a a+1<2极限存在，记为La→∞时L=√2a a=2例题3、a1>0a a+1=a(1+a a )a+a a(a>1)求极限lima→∞a a解：a a+1−a a=a(1+a a )a+a a −a(1+a a−1)a+a a−1=(a2−a)(a a−a a−1)(a+a a)(a+a a−1) a2−a1=a−a12a+a1当a1>√a a2−a1<0a a↓当0<a1≤√a a a↑所以0<a a+1=a(1+a a)a+a a <a(a+a a)a+a a=a极限存在a→∞时L=a(1+a)a+aa=√a 注：a a单调性有时依赖于a1的选取例题4、a1>1a a+1=11+a a 求极限lima→∞a a解：a a+1−a a=a a−1−a a(1+a a)(1+a a−1)（整体无单调性）a2a+1−a2a−1=11+a2a−11+a2a−2=a2a−2−a2a(1+a2a)(1+a2a−2) =a2a−1−a2a−3(1+a2a)(1+a2a−2)(1+a2a−1)(1+a2a−3)a3−a1=11+a2−a1<0所以{a2a+1}单调递减，同理，{a2a}单调递增有因为0<a a<1(a≥2)故lima→∞a2a+1和lima→∞a2a均存在，分别记为A,Ba2a+1=11+a2aa2a=11+a2a−1即A=11+B B=11+A解得 A=B=√5−1 2所以 lim a →∞a a =√5−12（7）泰勒公式法例题1、设f 有n 阶连续导数(a ≥2)a (a )(a 0)=0 (a =1,2,?,a −1)a (a )(a 0)≠0 ?a ∈aa (a 0+a )−a (a 0)=aa ′(a 0+aa ) (0<a =a (a )<1)证明：lim a →0a (a )=a11−a证明：a ′(a 0+aa )=a ′(a 0)+a "(a 0)(aa )+a 3(a 0)2!(aa )2+??+a (a −1)(a 0)(a −2)!(aa )a −2+a (a )(a )(a −1)!(aa )a −1即a ′(a 0+aa )=a (a )(a )(a −1)!(aa )a −1 a 0<a <a 0+aaa (a 0+a )= a (a 0)+a (a )(a )!a aa (a 0+a )−a (a 0)=a a a(a )(a )a !a 0<a <a 0+aa (a )(a )a !a a =a (a )(a )(a −1)!a a −1aa a −1θ=√a (a )(a )a (a )(a )a −11a 1a −1a →0 lim a →0a (a )=√a (a )(a )a (a )a a −1a 11−a = lim a →0a (a )=a 11−a（8）洛必达法则例题1、求lima→1a3−3a+2 a3−a2−a+1解：lima→1a3−3a+2a3−a2−a+1=lima→13a2−33a2−2a−1=lima→16a6a−2=32例题2、求lima→+∞a2−tan−1a1a解：lima→+∞a2−tan−1a1a=lima→+∞−11+x2−1a2=lima→+∞a21+a2=1例题3、求lima→+∞a aa aa(a为正整数，a>0)解：lima→+∞a aa aa=lima→+∞aa a−1aa aa=lima→+∞a(a−1)a a−2a2a aa=?=lima→+∞a!a a a aa=0例题4、求lima→0+a a ln a(a>0)解：lima→0+a a ln a=lima→0+ln aa−a=lima→0+1a−aa−a−1=lima→0+(−a aa)=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。