f(x)= [e^1/x -1]/ [e^1/x +1]当x→0时左右极限是什么,详细过程

2025-04-23 13:44:55
推荐回答(2个)
回答1:


回答2:

f(x)= [e^(1/x) -1]/ [e^(1/x) +1]

(x→0-)lim f(x)
= (x→0-) [e^(1/x) -1]/ [e^(1/x) +1]
= (0-1)/(0+1)
= -1

(x→0+)lim f(x)
= (x→0+) [e^(1/x) -1]/ [e^(1/x) +1]
= (x→0+) [1 - 1/e^(1/x)]/ [1 + 1/e^(1/x)]
= (1-0)/(1+0)
= 1