当x=0时 ∫" (0)=lim (0+h)-ln(1+0) h→0 f(0=lim ln[1+(0+h)l-ln(1+0) h→0 h f∫(0)=1 x<0 ∴f(x)= r>0 1+x当x = 0时, h h f h (0 ) ln(1 0) (0) lim 0 + − + = → − − = 1, h h f h ln[1 (0 )] ln(1 0) (0) lim 0 + + − + = + → + = 1, f (0) = 1. . , 0 1 1 1, 0 ( ) + = x x x f x