极限单元测试卷(满分:150分 时间:120分钟)一、选择题(本大题共12小题,每小题5分,共60分) 1.下面四个命题中,不正确...的是( ) A .若函数f (x )在x =x 0处连续,则lim x →x +0f (x )=lim x →x -0f (x )B .函数f (x )=x +2x 2-4的不连续点是x =2和x =-2C .若函数f (x )、g (x )满足lim x →∞[f (x )-g (x )]=0,则lim x →∞f (x )=lim x →∞g (x ) D.lim x →1x -1x -1=12答案:C 解析:A 中由连续的定义知函数f (x )在x =x 0处连续,一定有lim n →x +0f (x )=lim x →x -0f (x ),且还满足lim x →x +0f (x )=lim x →x -0f (x )=f (x 0),故A 对.B 中函数f (x )=x +2x 2-4在x =2和x =-2无定义,故不连续,B 对.C 中只有lim x →∞f (x ),lim x →∞g (x )存在时,才有lim x →∞f (x )=lim x →∞g (x ),否则不成立. D 中lim x →1x -1x -1=lim x →1 1x +1=12,故D 对.故选C. 2.下列命题中:①如果f (x )=13x ,那么lim x →∞f (x )=0 ②如果f (x )=1x,那么lim x →∞f (x )=0 ③如果f (x )=x 2+3xx +3,那么lim x →-3f (x )不存在 ④如果f (x )=⎩⎨⎧x (x ≥0)x +2 (x <0),那么lim x →0f (x )=0 其中错误命题的个数是( ) A .0 B .1 C .2 D .3 答案:D解析:②中x →-∞时无意义; ③中lim x →-3f (x )=lim x →-3x =-3; ④中左、右极限不相等.故选D.3.(2009·阳泉模拟)lim n →∞ 1+2+3+…+nn 2等于( ) A .2 B .1 C.12D .0答案:C解析:lim n →∞ 1+2+3+…+n n 2=lim n →∞ n +12n =lim n →∞ 1+1n 2=12.故选C. 4.已知函数f (x )=⎩⎪⎨⎪⎧x 2+2x -3x -1 (x >1)ax +1 (x ≤1)在点x =1处连续,则a 的值是( )A .2B .-2C .3D .-4 答案:C解析:∵lim x →1+f (x )=lim x →1+ x 2+2x -3x -1=lim x →1+(x +3)=4, 又lim x →1-f (x )=lim x →1- (ax +1)=a +1,f (1)=a +1. ∴要使f (x )在x =1处连续, 需lim x →1+f (x )=lim x →1-f (x )=f (1). 即a +1=4,∴a =3.故选C.5.已知函数f (x )=⎩⎪⎨⎪⎧x 2+2x +1 (x ≤0)ax +b (x >0)在x =0处连续且可导,则a 、b 的值依次为( )A .1,1B .2,1C .1,2D .2,2答案:B解析:由连续性知b =1;由可导性知a =2.选B.6.(2009·天津六县区联考)lim n →∞ C 02n +C 22n +C 42n +…+C 2n2n1-4n等于( ) A .-1 B .-12C .-14D .0答案:B解析:∵C 02n +C 22n +C 42n +…+C 2n 2n=12×22n =12×4n , ∴lim n →∞ C 02n +C 22n +C 42n +…+C 2n2n 1-4n =lim n →∞12×4n 1-4n=lim n →∞ 12(14)n -1=-12.故选B. 7.设f (n )=1+12+13+…+13n -1(n ∈N *),那么f (n +1)-f (n )等于( )A.13n +2B.13n +13n +1C.13n +1+13n +2D.13n +13n +1+13n +2答案:D解析:∵f (n )=1+12+13+…+13n -1∴f (n +1)=1+12+13+…+13n -1+13n +13n +1+13n +2∴f (n +1)-f (n )=13n +13n +1+13n +2.故选D.8.lim n →∞ (1n +1-2n +1+3n +1-…+2n -1n +1-2n n +1)的值为( ) A .-1 B .0C.12D .1答案:A解析:原式=lim n →∞(-1)×nn +1=-1.故选A. 9.设正数a ,b 满足lim x →2 (x 2+ax -b )=4,则lim n →∞ a n +1+ab n -1a n -1+2b n等于( ) A .0 B.14C.12D .1 答案:B解析:由lim x →2(x 2+ax -b )=4,即22+2a -b =4,得2a =b ,∴lim n →∞ a n +1+a ·b n -1a n -1+2·b n=lim n →∞a n +1+2n -1·a na n -1+2n +1·a n=lim n →∞ 12n +1+14·1a 12n +1·1a 2+1a =14.故选B. 10.数列{a n }中a 1=2,且a n =12(a n -1+3a n -1)(n ≥2),若lim n →∞a n 存在,则lim n →∞a n 等于( ) A. 3 B .- 3 C .±3 D. 6答案:A解析:∵a 1=2,a n =12(a n -1+3a n -1),则a n >0,∴lim n →∞a n ≥0,又lim n →∞a n =lim n →∞a n -1∴lim n →∞a n =12(lim n →∞a n -1+3lim n →∞a n -1), 解得:lim n →∞a n = 3. 11.若(1+5x 2)n 的展开式中各项系数之和是a n ,(2x 3+5)n 的展开式中各项的二项式系数之和为b n ,则lim n →∞ a n -2b n3a n +4b n的值为( ) A .-23 B .-12C.12D.13 答案:D解析:令x =1,得各项系数之和为a n =6n ,(2x 3+5)n 的展开式中各项的二项式系数之和为b n =2n ,∴lim n →∞a n -2b n 3a n +4b n =lim n →∞ 6n-2×2n3×6n +4×2n =lim n →∞ 1-2×(13)n3+4×(13)n =13. 12.数列{a n }中,有lim n →∞[(5n +2)a n ]=2,并有lim n →∞a n 存在,则lim n →∞ (na n )的值为( ) A .0B .2C.25D .不存在答案:C解析:因为lim n →∞a n 存在,可设lim n →∞a n =a , 又有lim n →∞[(5n +2)a n ]=2,且lim n →∞ 1n=0, ∴lim n →∞ 1n(5n +2)a n =0×2=0. 又lim n →∞ 1n (5n +2)a n =lim n →∞ (5a n+2a nn) =5lim n →∞a n+2lim n →∞ a nn =5a +0=0, ∴a =lim n →∞a n =0. ∴lim n →∞[(5n +2)a n ]=lim n →∞ (5na n +2a n ) =lim n →∞5na n +lim n →∞2a n =5lim n →∞na n +0=2. ∴lim n →∞ (na n)=25. 二、填空题(本大题共4小题,每小题5分,共20分)13.在数列{a n }中,a 1=9,且对任意大于1的正整数n ,点(a n ,a n -1)在直线x -y -3=0上,则lim n →∞ a n(n +1)2=________. 答案:9解析:由题意,得a n -a n -1=3,∴{a n }是等差数列. ∴a n =a 1+(n -1)×3=3n . ∴a n =9n 2.∴lim n →∞a n (n +1)2=lim n →∞ 9n 2(n +1)2=lim n →∞ 9(1+1n )2=9.14.lim x →π (x -π)cos xx -π=________.答案:-2π 解析:lim x →π(x -π)cos xx -π=lim x →π (x +π)cos x =(x +π)cos π=-2π. 15.如右图,连结△ABC 的各边中点得到一个新的△A 1B 1C 1,又连结△A 1B 1C 1的各边中点得到△A 2B 2C 2,如此无限继续下去,得到一系列三角形:△ABC 、△A 1B 1C 1、△A 2B 2C 2、…,这一系列三角形趋向于一个点M ,已知A (0,0),B (3,0),C (2,2),则点M 的坐标是 .答案:(53,23)解析:由条件结合图象可知,三角形的顶点都在△ABC 的三条中线上,由极限知识知M点的坐标是△ABC 的重心,∴(53,23)即为所求.16.将杨辉三角中的每一个数C r n 都换成分数1(n +1)C r n,就得到一个如图所示的分数三角形,称为莱布尼茨三角形.从莱布尼茨三角形可看出1(n +1)C r n +1(n +1)C x n =1nC r n -1,其中x =________.令a n =13+112+130+160+…+1nC 2n -1+1(n +1)C 2n,则lim n →∞a n =________. 11 12 12 13 16 13 14 112 112 14 15 120 130 120 15 16 130 160 160 130 16 17 142 1105 1140 1105 142 17答案:r +1,12解析:令n =3,14C r 3+14C x 3=13C r 2.当r =1时,14×3+14C x 3=13×2,14C x 3=16-112=112,∴C x 3=3.∴x =1,2.当r =2时,14C 23+14C x 3=13.∴14C x 3=13-112=312=14. ∴C x 3=1.∴x =3.归纳x =r +1. 利用裂项求和求极限求出lim x →∞a n 的值. 三、解答题(本大题共6小题,共70分)17.(本小题满分10分)已知f (x )=a ·b x (a 、b 为常数)的图象经过点P (1,18)和Q (4,8).(1)求f (x )的解析式;(2)记a n =log 2f (n ),n ∈N *,S n 是数列{a n }的前n 项和,求lim n →∞ S n3n 2+1. 解:(1)∵f (x )的图象经过点P (1,18)和Q (4,8),∴⎩⎪⎨⎪⎧ ab =18,ab 4=8,解得⎩⎪⎨⎪⎧a =132,b =4.∴f (x )=132×4x =4x -52=22x -5.(2)a n =log 2f (n )=log 222n -5=2n -5. ∵a n +1-a n =2(n +1)-5-(2n -5)=2, ∴{a n }是以-3为首项,公差为2的等差数列. ∴S n =n (-3+2n -5)2=n 2-4n .∴lim n →∞ S n3n 2+1=lim n →∞n 2-4n 3n 2+1=13. 18.(本小题满分12分)已知数列{a n }的前n 项和为S n ,其中a n =S n n (2n -1)且a 1=13.(1)求a 2、a 3;(2)猜想数列{a n }的通项公式,并用数学归纳法加以证明; (3)求lim n →∞S n . 解:(1)由a 2=13+a 22×3,得a 2=13×5,由a 1=11×3,a 2=13×5,得a 3=13+13×5+a 33×5,得a 3=15×7.(2)猜想a n =1(2n -1)(2n +1).证明:①当n =1时,显然成立.②假设n =k 时,猜想成立,即a k =1(2k -1)(2k +1),则n =k +1时,a k +1=S k +1(k +1)(2k +1),得S k +1=(k +1)(2k +1)a k +1,同时S k =k (2k -1)a k =k2k +1.两式相减,得a k +1=S k +1-S k =(k +1)(2k +1)a k +1-k 2k +1,即k (2k +3)a k +1=k2k +1.∴a k +1=1(2k +1)(2k +3),即n =k +1时,猜想成立.综上,a n =1(2n -1)(2n +1)(3)lim n →∞S n=lim n →∞[11×3+13×5+…+1(2n -1)(2n +1)] =lim n →∞ 12(1-13+13-15+…+12n -1-12n +1) =lim n →∞ 12(1-12n +1)=12. 19.(本小题满分12分)已知等比数列{a n }的首项为a 1,公比为q ,且有lim n →∞ (a 12+q -q n)=14,求首项a 1的取值范围.解:∵lim n →∞ (a 12+q -q n)=14, ∴0<|q |<1或q =1.当0<|q |<1时,即有0<|4a 1-2|<1.解之,得14<a 1<34,a 1≠12;当q =1时,lim n →∞ (a 13-1)=14, 即a 13-1=14,得a 1=154. 故a 1的取值范围为14<a 1<34且a 1≠12或a 1=154.20.(本小题满分12分)在边长为l 的等边△ABC 中 ,⊙O 1为△ABC 的内切圆,⊙O 2与⊙O 1外切,且与AB 、BC 相切,…,⊙O n +1与⊙O n 外切,且与AB 、BC 相切,如此无限继续下去,记⊙O n 的面积为a n (n ∈N *).(1)证明{an }是等比数列; (2)求lim n →∞(a 1+a 2+…+an )的值. (1)证明:记r n 为⊙O n 的半径,则r 1=l 2·tan30°=36l ,r n -1-r n r n -1+r n=sin30°=12,∴r n =13r n -1(n ≥2).于是a 1=πr 21=πl 212,a n a n -1=(r n r n -1)2=19,故{a n }成等比数列. (2)解:∵a n =(19)n -1a 1,∴lim n →∞ (a 1+a 2+…+a n )=a 11-19=3πl 232. 21.(本小题满分12分)已知公比为q (0<q <1)的无穷等比数列{a n }各项的和为9,无穷等比数列{a 2n }各项的和为815. (1)求数列{a n }的首项a 1和公比q ;(2)对给定的k (k =1,2,…,n ),设T (k )是首项为a k ,公差为2a k -1的等差数列,求数列T (2)的前10项之和;(3)设b i 为数列T (i )的第i 项,S n =b 1+b 2+…+b n ,求S n ,并求正整数m (m >1),使得lim n →∞ S nn m存在且不等于零. (注:无穷等比数列各项的和即当n →∞时该无穷等比数列前n 项和的极限)解:(1)依题意,可知⎩⎪⎨⎪⎧a 11-q=9a 211-q 2=815⇒⎩⎪⎨⎪⎧a 1=3,q =23.(2)由(1),知a n =3×(23)n -1,所以数列T (2)的首项为t 1=a 2=2,公差d =2a 2-1=3,S 10=10×2+12×10×9×3=155,即数列T (2)的前10项之和为155, (3)b i =a i +(i -1)(2a i -1) =(2i -1)a i -(i -1)=3(2i -1)(23)i -1-(i -1)S n =45-(18n +45)(23)n -n (n -1)2,lim n →∞S n n m =lim n →∞ (45n m -18n +45n m (23)n -n (n -1)2n m). 当m =2时,lim n →∞ S n n m =-12, 当m >2时,lim n →∞ S nn m =0,所以m =2. 22.(本小题满分12分)已知数列{a n }的首项a 1=5,前n 项和为S n ,且S n +1=2S n +n +5(n ∈N *).(1)证明数列{a n +1}是等比数列;(2)令f (x )=a 1x +a 2x 2+…+a n x n ,求函数f (x )在点x =1处的导数f ′(1),并比较2f ′(1)与23n 2-13n 的大小.(1)证明:由已知S n +1=2S n +n +5,∴n ≥2时,S n =2S n -1+n +4.两式相减,得S n +1-S n =2(S n -S n -1)+1,即a n +1=2a n +1. 从而a n +1+1=2(a n +1). 当n =1时,S 2=2S 1+1+5, ∴a 1+a 2=2a 1+6. 又a 1=5,∴a 2=11. 从而a 2+1=2(a 1+1).故总有a n +1+1=2(a n +1),n ∈N *. 又∵a 1=5,∴a n +1≠0. 从而a n +1+1a n +1=2,即{a n +1}是以a 1+1=6为首项,2为公比的等比数列. (2)解:由(1)知a n =3×2n -1. ∵f (x )=a 1x +a 2x 2+…+a n x n , ∴f ′(x )=a 1+2a 2x +…+na n x n -1, 从而f ′(1)=a 1+2a 2+…+na n=(3×2-1)+2(3×22-1)+…+n (3×2n -1) =3(2+2×22+…+n ×2n )-(1+2+…+n ) =3[n ×2n +1-(2+…+2n )]-n (n +1)2=3(n ×2n +1-2n +1+2)-n (n +1)2=3(n -1)·2n +1-n (n +1)2+6.由上,知2f ′(1)-(23n 2-13n ) =12(n -1)·2n -12(2n 2-n -1) =12(n -1)·2n -12(n -1)(2n +1) =12(n -1)[2n -(2n +1)].(*) 当n =1时,(*)式=0, ∴2f ′(1)=23n 2-13n ; 当n =2时,(*)式=-12<0, ∴2f ′(1)<23n 2-13n ; 当n ≥3时,n -1>0,又2n =(1+1)n =C 0n +C 1n +…+C n -1n +C n n ≥2n +2>2n +1,∴(n -1)[2n -(2n +1)]>0,即(*)>0,从而2f ′(1)>23n 2-13n . 〔或用数学归纳法:n ≥3时,猜想2f ′(1)>23n 2-13n . 由于n -1>0,只要证明2n >2n +1.事实上, ①当n =3时,23>2×3+1.不等式成立. ②设n =k 时(k ≥3),有2k >2k +1, 则2k +1>2(2k +1)=4k +2 =2(k +1)+1+(2k -1). ∵k ≥3,∴2k -1<0.从而,2k +1>2(k +1)+1+(2k -1)>2(k +1)+1,即n =k +1时,亦有2n >2n +1. 综合①②,知2n >2n +1对n ≥3,n ∈N *都成立.∴n ≥3时,有2f ′(1)>23n 2-13n .〕 综上,n =1时,2f ′(1)=23n 2-13n ; n =2时,2f ′(1)<23n 2-13n ; n ≥3时,2f ′(1)>23n 2-13n .。