3 2x+3 另解lm()=lim( 2x+1 1+ 2x 3 (1+)2lim( ×、J 2x x→0 2x = x→0 (1+)im(1+) 2x 2x e另解 = + + → x x x x ) 2 1 2 3 lim( x x x x x ) 2 1 (1 ) 2 3 (1 lim + + = → = e. 2 1 2 3 e e = x x x x ) 2 1 1 2 3 1 lim( + + → x x x x x x ) 2 1 lim(1 ) 2 3 lim(1 + + = → →