[求助] 考研数学求助，渐近线 已有3人参与

2楼: Originally posted by wurongjun at 2015-07-26 12:44:43
1.>> syms x
>> limit((1+exp(-x.^2))./(1-exp(-x.^2)),x,inf)
ans =
1
>> limit((1+exp(-x^2))/(1-exp(-x^2)),x,0)
ans =
inf
所以有两条渐近线:y=1和x=0
2.>> limit(exp(x),x,-i ...