5)因为 lim x[In(1+x)-lnx]=lmln(1+-)= In lim(1+-)=1,所以 lim n[In(1+n)-Inn]=1 lim(-- 1(-x)--x) (6) lim( lim( →x x→ 7.设函数 x>0 arcsin ax x<0 试求a、b,使∫(x)处处连续 解∫(x)处处连续,则必在x=0处连续,故 lim f(x)=lim 0x2f(0)=b,即b= limf(x)=Ⅷ m arcsin a ==f(0)=b,故a=2b=14 (5) 因为 1 1 lim [ln(1 ) ln ] lim ln(1 ) ln lim (1 ) 1 x x x xx x xx →∞ →∞ →∞ x x +− = + = + = , 所以 lim [ln(1 ) ln ] 1 n n nn →∞ + − = . (6) 1 1 cot lim( ) 1 1 cot ( )( ) tan 1 1 1 lim ( ) lim (1 ) e e x x x x x xx x x x x x →∞ − − − − →∞ →∞ − =− = = . 7. 设函数 1 1, 0, ( ) , 0, arcsin , 0. 2 x x x fx b x ax x x ⎧ + − ⎪ > ⎪⎪ = ⎨ = ⎪ ⎪ < ⎪⎩ 试求 a 、b , 使 f ( ) x 处处连续. 解 () f x 处处连续, 则必在 x = 0 处连续, 故 00 0 1 11 1 2 lim ( ) lim lim (0) xx x 2 x x f x fb x x →→ → ++ + + − = = == = , 即 1 2 b = ; 0 0 arcsin lim ( ) lim (0) x x 2 2 ax a f x fb x → → − − = == = , 故 a b = 2 1 =