2.lim(1+)=e x→00 证:当x>0时,设n≤x<n+1,则 (1+)”<1+)<1+)+ im0+)”=Jim -=e n->0 n-→0 1+ n+1 0+-m+y1+]=e n-→00 lim(1+)'=e X→+00 OOo⊙⊙8 机 2. 证: 当 x 0 时, 设 n x n +1, 则 x x (1 ) 1 + 1 1 (1 ) + + n n + + n n (1 ) 1 1 n n n lim (1 ) 1 1 + → + lim → = n 1 1 1 (1 ) + + + n n 1 1 1 + + n = e 1 1 lim (1 ) + → + n n n lim[(1 ) 1 ] 1 n n( 1 n ) n = + + → = e e x x x + = →+ lim (1 ) 1 机动 目录 上页 下页 返回 结束