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 < ⋅ + ⋅ ≤⋅ +⋅ = ε . 因此 lim[ ( ) ( )] lim ( ) lim ( ) x xx f x g x AB f x g x →∞ →∞ →∞ ⋅ == ⋅ . (3) 因为 () 1 lim lim[ ( ) ] x x () () f x f x →∞ →∞ g x gx = ⋅ , 故由(2), 只需证当 B ≠ 0 时, 有 11 1 lim x ( ) lim ( ) x →∞ g x B gx →∞ = = . 1 1 () 1 1 ( ) () () () B gx g x B gx B gx B B gx − −= =⋅ ⋅ − ⋅ . 由 lim ( ) x g x B →∞ = 及函数极限的局部有界性知, ∀ε > 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 →∞ = =