f∫(x 证:(0)=lim p(x)-p(0) f(0) f(x)-xf(0) x→>0 x→>0 x-0 2 li f(x)-f'(0) lim f"(x) f" (0 x→>0 2x x→02 2 x(x)-∫(x) x≠0 ∴p(x)= x=0 xf"(x)+f'(x)-f( ∴limp(x)=lim xf(x)-f(x) x→>0 x→>0 2x lim f"(x)f"(0) x→02 2 p(0).→d(x)有一阶连续导数x x x ( ) (0) (0) lim 0    −  = → 证： 2 0 0 ( ) (0) lim (0) ( ) lim x f x xf x f x f x x x −  = −  = → → x f x f x 2 ( ) (0) lim 0  −  = → 2 ( ) lim 0 f x x  = → . 2 f (0) =        =    −   = , 0 2 (0) , 0 ( ) ( ) ( ) 2 x f x x xf x f x  x 2 0 0 ( ) ( ) lim ( ) lim x xf x f x x x x  −   = → →  x xf x f x f x x 2 ( ) ( ) ( ) lim 0  +  −  = → (0). 2 (0) 2 ( ) lim 0 =   =  = → f x f x (x)有一阶连续导数