来自郭丁的问题
【lim(2^x/1次幂-1)/(2^x/1次幂+1)】
lim(2^x/1次幂-1)/(2^x/1次幂+1)
1回答
2020-04-11 12:36
【lim(2^x/1次幂-1)/(2^x/1次幂+1)】
lim(2^x/1次幂-1)/(2^x/1次幂+1)
lim(x→0)[(2^x-1)/(2^x+1)]
=lim(x→0)(2^x-1)/[lim(x→0)(2^x+1)]
=(2^0-1)/(2^0+1)
=(1-1)/(1+1)
=0