2003数学二.pdf

2003 6 4 24 . 1 1 x →0 (1−ax 2 ) 4 −1 x sin x a= . -4 1 x →0 2 4 1 2 2 (1−ax ) −1~ − ax x sin x ~ x . 4 1 1 2 2 4 − ax (1−ax ) 4 1 lim lim − a 1 2 x →0 x sin x x →0 x 4 a=-4. 2 y=f(x)xy +2 ln x y 4 y=f(x)(1,1) . x-y=0 xy +2 ln x y 4 x ′ 2 3 ′ y +xy + 4y y x ′ x=1,y=1 y (1) 1. (1,1) y −1 1⋅(x −1) x −y 0. 3 y 2x x n ______. (ln 2)n n! y ′ 2x ln 2 y ′′ 2x (ln 2)2 ,y (x ) 2x (ln 2)n y (n) (0) (ln 2)n x n (n) n y (0) (ln 2) . n! n! aθ θ 4 ρ e (a 0) 0 2π ______. 1 4πa (e −1) 4a 1 2π 2 1 2π 2aθ S ∫ ρ (θ)dθ ∫ e dθ 2 0 2 0 1 2aθ 2π 1 4πa = e (e −1)