当x>0时，令x=-(t+1),则t→+0，从而有 1md+》=lm0-)0 t+00 =m品t切=m0+y =lim[1+'(1+】=e 故1im(+)=e X>0 说明：此极限也可写为lim(1+x)=e X0 BEIJING UNIVERSITY OF POSTS AND TELECOMMUNICATIONS PRESS 目录上页下页返回结束目录 上页 下页 返回 结束 当 x = −(t +1), 则 从而有 ( 1) 1 1 lim (1 ) − + + →+ = − t t t ( 1) 1 lim ( ) − + + →+ = t t t t 1 1 lim (1 ) + →+ = + t t t lim [(1 ) (1 )] 1 1 t t t t = + + →+ = e 故 lim (1 ) e 1 + = → x x x 说明: 此极限也可写为 lim (1 ) e 1 0 + = → x x x 时, 令