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)-4<5;当>x2时,有g(x)-B<5 取X=max{x1,X2},则当x>X时,|f(x)+g(x)-(A+Bs|(x)-4 +g(x)-B<5+=E, Elt lim[(x)+g(x)]=A+B= lim f(x)+lim g(x) (2)由于f(x)g(x)-ABl=|(x)·g(x)-Bf(x)+B(x)-AB ≤(x)g(x)-B+|B|(x)-4 由limf(x)=A及函数极限的局部有界性得,VE>0,彐x1>0及M>0,当 x>X1时,有|(x)-4<,且f(x)≤M,其中C=max{M,B 又mg(x=B,故3X2>0,当>x2时,有g(x)-B<2C3 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 − < ; 当 2 x > X 时, 有 ( ) 2 gx B ε − < . 取 X XX = max{ , } 1 2 , 则当 x > X 时, [ ( ) ( )] ( ) f x gx A B + −+ ≤ f ( ) x A − ( ) 2 2 gx B ε ε + − <+= ε , 因此 lim[ ( ) ( )] lim ( ) lim ( ) x xx f x gx A B f x gx →∞ →∞ →∞ + =+= + . (2) 由于 f () () () () () () x g x AB f x g x Bf x Bf x AB ⋅ −= ⋅ − + − ≤ f () () x gx B ⋅ − + B fx A ⋅ − ( ) . 由 lim ( ) x f x A →∞ = 及函数极限的局部有界性得, ∀ε > 0 , 1 ∃ X > 0 及 M > 0 , 当 1 x > X 时, 有 ( ) 2 fx A C ε − < , 且 f ( ) x M≤ , 其中C MB = max{ , }. 又 lim ( ) x g x B →∞ = , 故 2 ∃ > X 0 , 当 2 x > X 时, 有 ( ) 2 gx B C ε − <