第五节极限的运算法则 习题 1.下列运算是否正确,为什么? (1)lim(-+ n→∞nn+1 n+n =0+0+…+0=0 ()m(kx+-√x-1)= x+1-lin (3) lim xsin -= limx. lim\t0 解(1)不正确,因为只有有限个数列和的极限(且这有限个数列的极限都存 )才等于它们极限的和 (2)不正确,因为只有当两函数极限都存在时,才有两函数差的极限等于它们 极限的差 (3)不正确,因为msn不存在 2.计算下列各极限 (2)lim(1 ); (3) lim n3-n+3 n→∞122.3 n(n+1) m2+1-m2-2n) (n+1)(2n+1)(3n+1) 解(1)lim 1+2+3 (n-1) n→① (2)lm(1+++…+一)=lim →, =m21-(2y")
1 第五节 极限的运算法则 习 题 1-5 1. 下列运算是否正确, 为什么? (1) 11 1 lim ( ) n→∞ n n nn 1 + ++ + + " 11 1 lim lim lim nn n →∞ →∞ →∞ n n nn 1 = + ++ + + " =++ += 00 00 " ; (2) lim ( 1 1) lim 1 lim 1 0 x xx xx x x →+∞ →+∞ →+∞ + − − = + − − =∞−∞= ; (3) 0 00 1 1 lim sin lim lim sin 0 x xx x x → →→ x x =⋅ = . 解 (1) 不正确, 因为只有有限个数列和的极限(且这有限个数列的极限都存 在)才等于它们极限的和. (2) 不正确, 因为只有当两函数极限都存在时, 才有两函数差的极限等于它们 极限的差. (3) 不正确, 因为 0 1 lim sin x→ x 不存在. 2. 计算下列各极限: (1) 2 1 2 3 ( 1) lim n n →∞ n + ++ + − " ; (2) 11 1 lim (1 ) 2 4 2n n→∞ +++ + " ; (3) 2 3 5 23 lim n 3 n n →∞ n n + + − + ; (4) 11 1 lim ( ) n→∞ 1 2 2 3 ( 1) n n + ++ ⋅ ⋅ + " ; (5) 2 2 lim ( 1 2 ) n n nn →∞ +− − ; (6) 3 ( 1)(2 1)(3 1) lim n 3 nnn →∞ n + + + . 解 (1) 2 1 2 3 ( 1) lim n n →∞ n +++ + − " 2 ( 1) 2 1 lim n 2 n n →∞ n − = = . (2) 1 1 1 1 (1 ( ) ) 11 1 1 2 lim (1 ) lim lim 2(1 ( ) ) 2 24 2 2 1 1 2 n n n n nn + + →∞ →∞ →∞ ⋅ − + ++ + = = − = − "
5n2+2n+3 lm(1-2+) 月→① 11 o)=lim(1 n m(1--) (5) lim( (6)lm(n+1X2n+1)3n+1 (1+-)(2+-)(3+- →① 3 3.计算下列各极限 3x2-7x+1 lim (4) lim (5)im(1+-)2-2) (6) lim- 解(1)lim(1--) 25 3x2-7x+1 3 lim(5+-- (3)im =-lim( (x2+x+1)=3 (5)ma1+-2、1 )=(1+lim)(2-lim
2 (3) 2 23 23 3 23 23 52 3 52 3 lim ( ) 5 23 lim lim 0 3 13 13 1 lim (1 ) n n n n n n n n nn nn n n nn nn →∞ →∞ →∞ →∞ ++ ++ + + == = − + −+ −+ . (4) 1 1 1 111 1 1 lim ( ) lim (1 ) n n →∞ →∞ 1 2 2 3 ( 1) 2 2 3 1 nn n n + + + = −+−+ +− ⋅ ⋅+ + " " 1 lim (1 ) n→∞ n 1 = − + =1. (5) 2 2 2 2 2 1 2 1 2 lim ( 1 2 ) lim lim 1 2 1 2 1 1 n nn n n n nn n nn n n →∞ →∞ →∞ + + +− − = = ++ − + + − =1. (6) 3 111 (1 )(2 )(3 ) ( 1)(2 1)(3 1) lim lim 2 n n 3 3 nnn nnn →∞ →∞ n +++ +++ = = . 3. 计算下列各极限: (1) 0 2 lim(1 ) x→ x 3 − − ; (2) 2 2 3 71 lim x 5 23 x x →∞ x x − + + − ; (3) 2 0 2 lim 1 1 x x x → − + ; (4) 3 1 1 lim x 1 x → x − − ; (5) 2 1 1 lim (1 )(2 ) x→∞ x x + − ; (6) 2 2 1 1 lim x 2 1 x → x x − − − . 解 (1) 0 0 2 2 25 lim(1 ) 1 lim 1 x x → → x x 3 3 33 − =− =− = − −− . (2) 2 2 2 2 2 2 71 71 3 lim (3 ) 3 71 3 lim lim 5 23 23 23 5 5 lim (5 ) x x x x x x x x x x x x x x x x →∞ →∞ →∞ →∞ −+ −+ − + == = + − +− +− . (3) 2 2 0 0 2 lim lim(1 1 ) 2 1 1 x x x x x → → =− + + =− − + . (4) 3 2 1 1 1 lim lim( 1) 3 x x 1 x x x → → x − = ++ = − . (5) 2 2 11 1 1 lim (1 )(2 ) (1 lim )(2 lim ) 2 x xx →∞ →∞ →∞ x x x x + − =+ − = . (6) 2 2 1 1 1 12 lim lim x x 2 1 2 13 x x → → x x x − + = = − − +
4.计算下列各极限 (1)im(x+)2-x2 (2)1m(1-2)1-2)…(1-) h 解(1)lm(x+h)2-x =lm(2x+h)=2 h→0 (2)im(1-(1-5)…(1-2)=lm(1-(1+-(1+÷)…(1--)1+-) =lim(1+-)= 5.表述并证明x→∞时函数极限的四则运算法则 解若limf(x)=A,limg(x)=B,则 x→① (1)Iim[f(x)±g(x)]=A±B=limf(x)±limg(x) (2) lim(f(x).g(x)]=AB= lim f(x). lim g(x) x→① B≠0,则有limf(x)4Imf(x) B lim g(x) 证明如下: (1)仅证明和的形式 由limf(x)=A,lm8(x)=B知,ⅤE>0,彐X1>0.,X2>0,当x>X1 时,有(x)-4x2时,有g(x)-BX时,|f(x)+g(x)-(A+Bs|(x)-4 +g(x)-B0,彐x1>0及M>0,当 x>X1时,有|(x)-40,当>x2时,有g(x)-B<2C
3 4. 计算下列各极限: (1) 2 2 0 ( ) lim h x h x → h + − ; (2) 22 2 11 1 lim (1 )(1 ) (1 ) n→∞ 2 3 n −− − " . 解 (1) 2 2 0 0 ( ) lim lim(2 ) 2 h h xh x x h x → → h + − = += . (2) 22 2 1 1 1 1111 11 lim (1 )(1 ) (1 ) lim (1 )(1 )(1 )(1 ) (1 )(1 ) n n →∞ →∞ 2 3 n 2233 n n − − −= − + −+ − + " " 1 11 lim (1 ) n→∞ 2 2 n = + = . 5. 表述并证明 x → ∞ 时函数极限的四则运算法则. 解 若 lim ( ) x f x A →∞ = , lim ( ) x g x B →∞ = , 则 (1) lim[ ( ) ( )] lim ( ) lim ( ) x xx f x gx A B f x gx →∞ →∞ →∞ ± =±= ± ; (2) lim[ ( ) ( )] lim ( ) lim ( ) x xx f x g x AB f x g x →∞ →∞ →∞ ⋅ == ⋅ ; (3) 若 B ≠ 0 , 则有 lim ( ) ( ) lim ( ) lim ( ) x x x f x fx A g x B gx →∞ →∞ →∞ = = . 证明如下: (1) 仅证明和的形式. 由 lim ( ) x f x A →∞ = , lim ( ) x g x B →∞ = 知, ∀ε > 0 , 1 2 ∃ X X > > 0, 0 , 当 1 x > X 时, 有 ( ) 2 fx A ε − X 时, 有 ( ) 2 gx B ε − X 时, [ ( ) ( )] ( ) f x gx A B + −+ ≤ f ( ) x A − ( ) 2 2 gx B ε ε + − 0 , 1 ∃ X > 0 及 M > 0 , 当 1 x > X 时, 有 ( ) 2 fx A C ε − X 0 , 当 2 x > X 时, 有 ( ) 2 gx B C ε − <
取X=max{x1,X2},则当>X时,有 (x):g(x)-AB|s|(x)g(x)-B+1B|(x)-4 因此回m(x)8(x)=AB=mnf(x)img(x) (3)因为m(x=1im(x)-1-1,故由(2)只需证当B≠0时,有 x→g(x)x→x x)B lim g(x) B-g(x)_1 g(x) g(x).BB lg(x) (x)-B 由limg(x)=B及函数极限的局部有界性知,VE>0,彐X>0及M>0,当 X时,有g()-2,且≤M,所以 B B 8(xl 8(x)-B B mg(x)
4 取 X XX = max{ , } 1 2 , 则当 x > X 时, 有 f () () x g x AB ⋅ − ≤ ⋅− f () () x gx B + B fx A ⋅ − ( ) 2 222 M B CC C CCC ε εεε 0 , 0 ∃ X > 及 M > 0 , 当 x > X 时, 有 ( ) B gx B M − < ε , 且 1 ( ) M g x ≤ , 所以, 1 111 1 ( ) () () B gx B M gx B B gx B M − = ⋅ ⋅ −< ⋅⋅ = ε ε . 即 11 1 lim x ( ) lim ( ) x →∞ g x B gx →∞ = =