(2^x-1)/x当x趋近于0时的极限怎么求?

【罗必塔法则】
lim(x->0) (2^x-1)/x
=lim(x->0) ln2 * 2^x /1
= ln2
【等价无穷小量】
令：2^x - 1 = t, 则：x = ln(1+t)/ln2 , x->0 ,t->0 ,ln(1+t)~ t
lim(x->0) (2^x-1)/x
=lim(x->0)t/[ln(1+t)/ln2]
=lim(x->0)ln2 t/ln(1+t)
= 1
【重要极限】
令：2^x - 1 = t, 则：x = ln(1+t)/ln2 , x->0 ,t->0
lim(x->0) (2^x-1)/x
=lim(x->0)t/[ln(1+t)/ln2]
=lim(x->0)ln2/ln[(1+t)^(1/t)]
= ln2/lne
= ln2