(3) lim( (4) lim (5) lim(1+3tan'x) (7)lim(+a(1+a2)…(+a)l<1; (8)lim(x+-+…+ 2 cos 解(1) lim sin x-sna=lm x+a Im cos x-a (2)lim(√x2 1-√x2-2x+3)=lim x-1+√x2 4 lim (3)1m(,2 lim( x→!1+x +xsinx-l sinx (4) lim (5) lim(+ 3 tan x)ot=[lim(1+3 tan x)tanP=e (6)设k为任一个大于2c的自然数,则当n>k时 0<=(12kk+1k+2y=2y 由于m(2:0,由夹遥准则,〃O 7)lim(1+a+a2)…(+a2")=lim n(1-a)(1+a)1+a2)…(1+a2) im(-a-)=2 (3) 1 1 2 1 1 lim( ) 1 x x x x x − − → − − ; (4) 2 0 1 sin 1 lim e 1 x x x x → + − − ; (5) 2 2 cot 0 lim(1 3tan ) x x x → + ; (6) lim ( 0) ! n n c c →∞ n > ; (7) 2 2 lim (1 )(1 ) (1 ), 1 n n a a aa →∞ + ++< " ; (8) 1 2 lim ( ) n 2! 3! ( 1)! n →∞ n +++ + " . 解 (1) 2cos sin sin sin sin 22 2 lim lim lim cos lim 2 2 xa xa xa xa x a xa xa x a xa → → →→ xa xa x a + − − ⋅ − + = =⋅ − − − cos 1 cos 2 a a a + = ⋅= . (2) 2 2 2 2 3 4 lim ( 1 2 3) lim 1 23 x x x xx x x xx x x →+∞ →+∞ − + −− − + = + −+ − + 2 2 4 3 3 lim 11 23 2 1 1 x x x x x x →+∞ − = = +− + −+ . (3) 1 1 1 1 1 2 2 1 1 1 1 11 lim( ) lim( ) ( ) 1 1 2 2 x x x x x x x x − − + → → − = == − + . (4) 2 2 0 0 1 sin 1 sin 1 1 2 lim lim e 1 2 x x x x x x x → → x + − = = − . (5) 2 2 1 2 cot 2 3 3 3tan 0 0 lim(1 3tan ) [lim(1 3tan ) ] e x x x x x x → → + =+ = . (6) 设 k 为任一个大于 2c 的自然数, 则当 n k > 时, 1 (2 ) 0 ( )( ) ( ) ! 12 1 2 2 2 n k k nk n c cc c c c c c c n kk k n − < =⋅ ⋅ <⋅ = + + " " , 由于 (2 ) lim 0 2 k n n c →∞ = , 由夹逼准则, 故 lim 0 ! n n c →∞ n = . (7) 22 22 1 lim (1 )(1 ) (1 ) lim (1 )(1 )(1 ) (1 ) 1 n n n n aa a aaa a →∞ →∞ a ++ + = −++ + − " " 1 1 1 2 lim(1 ) 1 1 n n a a a + →∞ = −= − −