xe dx=xe d xe dx (1+x) 1+x1+xJ1+x 1+ =-(x-1)ex+C=°,+C,又g(0)=1,故C=0 1+ 于是g(x)=1,。所以 2(1+x 提问解答: 计算lm43 解:lm∏ k-1(k2+k+1) lim lim( n∞2k3+1m→∞d2(k+1k2-k+1)m→∞k2k+1 lim( lim n2 2 (1 ) 1 1 1 x x xe x x x e x x dx xe d dxe x x x x = = − + + + + ∫ ∫ ∫ = 2 1 x x e x xe dx x − + ∫ = 2 ( 1) 1 x x e x x e C x − − + + = 1 x e C x + + ，又 g(0) =1，故C = 0。 于是 ( ) 1 x e g x x = + 。所以 2 ( ) 2(1 ) 1 x x xe f x e x x = + + 。 提问解答： 计算 3 3 2 1 lim 1 n n k k →∞ k = − + ∏ 。 解： 3 3 2 1 lim 1 n n k k →∞ k = − + ∏ = 2 2 2 2 2 2 ( 1)( 1) 1 1 lim lim ( ) ( 1)( 1) 1 1 n n n n k k k k k k k k k →∞ k k k →∞ k k k = = 2 n = − + + − + + = ⋅ + − + + − + ∏ ∏ ∏ = 2 2 1 2 3 1 7 13 1 lim ( )( ) n 345 1 3 7 1 n n n + →∞ n n n − + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + − + " " 2 1 2 1 lim n ( 1) 3 n n →∞ n n ⋅ + + ⋅ + = 2 3 = 。 2