lim(x→0)(ln(1+2x)/(e^x-1))

问题描述:

lim(x→0)(ln(1+2x)/(e^x-1))

x→0时,1+2x~2x,e^x-1~x(等价无穷小)
所以
lim(x→0)(ln(1+2x)/(e^x-1))
=lim(x→0)(2x/x)
=2

lim(x→0)(ln(1+2x)/(e^x-1))
=lim(x→0)[(2/(1+2x)e^x] .......分子分母求导
=2

0/0型极限,用L'Hospital法则
lim(x→0)(ln(1+2x)/(e^x-1))
=lim(x→0)((ln(1+2x))'/(e^x-1)')
=lim(x→0)(2/(1+2x)/e^x)
=2

等于2啊,用洛比达法则,上下同求一次导数就行