第六节极限的存在准则与两个重要极限 习题 1.计算下列极限 (1) lim inax (2)Iim√xcot I-0 tan Bx 3)lim 3"sin I 1-cos 2x (4) lim (5)Im OSx x→∞2x+cosX 解(1)若a≠0,li lim snax os Bx= lim inax. Br.axa x-0 tan Bx I-0 sin Bx x-o ax sin BxBx B 若a=0,易知lim sinai (2) lim vxcot - A cos x= lin s√x=1 (4) lim √2 cOS x (6)lim x→∞2x+ costs→,COSx 2.计算下列极限: (1)lim(1+ax)x(a,b>0); (2)lim() x+1
1 第六节 极限的存在准则与两个重要极限 习 题 1-6 1. 计算下列极限: (1) 0 sin lim ( 0) x tan x x α β → β ≠ ; (2) 0 lim cot x x x → ＋ ; (3) π lim 3 sin 3 n n n→∞ ; (4) 0 1 cos 2 lim x sin x → x x − ; (5) 0 lim x 1 cos x x → + − ; (6) sin lim x 2 cos x x →∞ x x − + . 解 (1) 若α ≠ 0 , 00 0 sin sin sin lim lim cos lim xx x tan sin sin x x xxx x x x x xx α α α β αα β →→ → β β α ββ β = ⋅ = ⋅ ⋅= ; 若α = 0 , 易知 0 sin lim 0 x tan x x α α → β β = = . (2) 0 0 00 lim cot lim cos lim lim cos 1 x x xx sin sin x x xx x x → → →→ x x = ⋅= ⋅ = ＋ ＋ ＋＋ . (3) π sin π 3 lim 3 sin lim π π 3 π 3 n n n n n n →∞ →∞ = ⋅= . (4) 2 0 0 1 cos 2 2sin lim lim 2 x x sin sin x x → → xx xx − = = . (5) 0 00 2 lim lim lim 2 2 1 cos 2 sin sin 2 2 x xx x x x x x x → →→ + ++ = = ⋅= − . (6) sin 1 sin 1 lim lim 2 cos 2 cos 2 x x x x x x x x x x →∞ →∞ − − = =− + + . 2. 计算下列极限： (1) 0 lim(1 ) ( , 0) b x x ax a b → + > ; (2) 1 lim ( ) 1 x x x →∞ x − + ;
(3)lim1-2x (4) lim(1+cosx)- (5)Iim(1--)n(k为正整数) (6)lim() A2(1)lim(1+ ax) =lim(+ax)ar =eab (2) lim( x→∞x+1 x+1 (3)1m-2x=lm(-2x)2x2)=c2 (4) lim(1+cos x)sc= lim(1+ cosx)cost =e (5) lim( --)=lim[(1-)=e (6)lim()=lim(1+ 3.利用夹逼准则证明下列极限 (1) lim( 四吗+1+2++n)=2 (4)lim1+x=1 1)因为 n n l√m2+2 n2 又lim lim =1; lim-n n→, 所以im(
2 (3) 0 lim 1 2 x x x → − ; (4) 2sec π 2 lim(1 cos ) x x x → + ; (5) 1 lim (1 )k n n→∞ n − ( k 为正整数); (6) 1 lim ( ) 1 n n n →∞ n + − . 解 (1) 1 0 0 lim(1 ) lim(1 ) e b ab x ax ab x x ax ax ⋅ → → += + = . (2) 1 2 ( )( ) 1 2 2 1 2 lim ( ) lim (1 ) e 1 1 x x x x x x x x x + − ⋅− + − →∞ →∞ − =− = + + . (3) 1 ( )( 2) 2 2 0 0 lim 1 2 lim(1 2 ) e x x x x x x − − − → → −= − = . (4) 2 2sec 2 cos π π 2 2 lim(1 cos ) lim(1 cos ) e x x x x x x → → + =+ = . (5) 1 1 lim (1 ) lim[(1 ) ] e kn n k k n n n n − − − →∞ →∞ −= − = . (6) 1 2 1 2 2 1 2 lim ( ) lim (1 ) e 1 1 n n n n n n n n n − ⋅ − →∞ →∞ + =+ = − − . 3. 利用夹逼准则证明下列极限: (1) 22 2 11 1 lim ( ) 1 1 2 n n n nn →∞ + ++ = ++ + " ; (2) 22 2 12 1 lim ( ) n 1 2 2 n →∞ n n nn + ++ = ++ + " ; (3) 22 2 ππ π lim (sin sin sin ) π 1 2 n n n nn →∞ + ++ = ++ + " ; (4) 0 lim 1 1 n x x → + = . 证 (1) 因为 2 22 2 2 11 1 12 1 n n nn n n nn n < + ++ < + ++ + + " , 又 2 1 lim lim 1 1 1 n n n n n n →∞ →∞ = = + + ; 2 2 1 lim lim 1 1 1 1 n n n n n →∞ →∞ = = + + ; 所以 22 2 11 1 lim ( ) 1 1 2 n n n nn →∞ + ++ = ++ + "
n(n+1) (2)因为 n n2+nn2+1 n(n+1) -n(nt -n(n+ 1) 月→n+n lim (3)因为 ≤ √n2+1 又 lim nsin lim-vn2 =π,同理 lim nsir lim(sin +sin +…+Sin n2+1 n2+2 (4)当x>0时,1<√1+x<1+x,故lmh+x=1 当-1<x<0时,1+x<h+x<1,故imy1+x=1 故 lim√+x=1 4.利用单调有界准则证明下面数列存在极限,并求其极限值: √2…,a=√2(m次复合 (2)x=1,x2=1+-x 高十1”=1+如 证(1)易知an=√2an(m=12…),下证此数列单调有界 √2<2,假设n=k时 2a<2,即an<2(n=12…),即此数列有界
3 (2) 因为 2 2 2 2 22 2 1 ( 1) 2 12 12 1 2 n n n n n n n nn n n nn n n n + = + ++ 0 时, 11 1 n < + <+ x x , 故 0 lim 1 1 n x x → + + = ; 当 −< < 1 0 x 时, 1 11 n +< +< x x , 故 0 lim 1 1 n x x → − + = . 故 0 lim 1 1 n x x → + = . 4. 利用单调有界准则证明下面数列存在极限, 并求其极限值: (1) 1 2 2, 2 2, , 2 2 2 n aa a == = " " ( n 次复合); (2) 1 1 1 2 1 1 1, 1 , , 1 1 1 n n n x x xx x x x − − = =+ =+ + + " . 证 (1) 易知 1 2 ( 1,2, ) n n a an + = = " , 下证此数列单调有界: 当 n =1时, 1 a = < 2 2 , 假设 n k = 时, 2 k a < , 则当 n k = +1时, k 1 a + = 2 k a < 2 , 即 2( 1,2, ) n a n < = " , 即此数列有界;
因为 1+1> an 综上,iman存在,令iman=A 又an1=√2an,故an=2an,因此lan=2lman,即2=2A 解得4=2,A2=0(舍去),故iman=2 (2)易知xn>0,先证此数列单调有界: 当n=1时,x=1≤2,当n>1时,x=1+—m,≤2,即xn≤2(m=1,2,…),即 此数列有界; 0,故 x -x )-(1 xn-1+1(xn+1xn-1+1) >0 (xn+1)xn-1+1)…(x2+1)x+1) 综上,imxn存在,令imxn=A 因此imxn=1+ lim x,+ 解得A= A (舍去),故lmxn 57…(2n-1),(2n)!!=2.4.6.8…(2n) 设 (2n-1)!! ,试证明 并求极限lm (2n)! 分<xn< 证易知 (2(n+1)-1)!2n+1 (2(n+1)!! 当n=1时 假设n=k时 则当 =k+1时
4 因为 2 1 2 ( 2) 2 2 2 n n nn n n nn nn nn a a aa a a aa aa aa + − − − −= −= = + + , 由 2 n a , 即 n n 1 a a + > . 综上, lim n n a →∞ 存在, 令 lim n n a A →∞ = . 又 1 2 n n a a + = , 故 2 1 2 n n a a + = , 因此 2 1 lim 2 lim n n n n a a + →∞ →∞ = , 即 2 A = 2A , 解得 1A = 2 , 2 A = 0 (舍去), 故 lim 2 n n a →∞ = . (2) 易知 0 n x > , 先证此数列单调有界: 当 n =1时, 1x = ≤1 2 , 当 n >1时, 1 1 1 2 1 n n n x x x − − = + ≤ + , 即 2( 1,2, ) n x n ≤ = " , 即 此数列有界; 又 2 1 1 0 2 x x − => , 故 1 1 1 1 1 (1 ) (1 ) 1 1 ( 1)( 1) n n nn n n n n nn x x xx x x x x xx − − + − − − − =+ −+ = + + ++ 2 1 1 21 0 ( 1)( 1) ( 1)( 1) n n x x xx xx − − == > + + ++ " " , 即 n n 1 x x + > . 综上, lim n n x →∞ 存在, 令 lim n n x A →∞ = . 又 1 1 1 1 n n n x x x − − = + + , 因此 1 1 lim lim 1 lim 1 n n n n n n x x x − →∞ →∞ − →∞ = + + , 即 1 1 A A A = + + , 解得 1 1 5 2 A + = , 1 1 5 2 A − = (舍去), 故 1 5 lim 2 n n x →∞ + = . 5. 记(2 1)!! 1 3 5 7 (2 1) n n − =⋅⋅⋅ − " , (2 )!! 2 4 6 8 (2 ) n n = ⋅⋅⋅ " . 设 (2 1)!! (2 )!! n n x n − = , 试证明 1 1 4 21 n x n n < < + , 并求极限 lim n n x →∞ . 证 易知 1 (2( 1) 1)!! 2 1 (2( 1))!! 2 2 n n n n x x n n + +− + = =⋅ + + , 当 n =1 时, 1 1 11 4 21 2 < =< x + , 假设 n k = 时, 1 1 4 21 k x k k < < + , 则当 n k = +1时
k+12k+11 ④4k+12k+2√442k+242k+2k+1√k+D+1 故 由lim √4nn√2n+1 0,故
5 1 1 211 21 21 1 1 4( 1) 4 2 1 2( 1) 1 22 22 22 k k k kk x x k k kk k kk + + ++ < ⋅ < = ⋅< ⋅ < + + ++ + ++ , 故 1 1 4 21 n x n n < < + . 由 1 1 lim lim 0 n n →∞ →∞ 4 21 n n = = + , 故 lim 0 n n x →∞ =
