Question

大一高数题，求右极限的，求大神帮忙解答，感激不尽～

Answer 1

原式=lim(x→0+)[1+(cos√x-1)]^(π/x)
=lim(x→0+)(1-x/2)^(π/x)
=lim(x→0+)[(1-x/2)^(-2/x)]^(-π/2)
=e^(-π/2)

追问

亲，第一步到第二步原因是什么呢？

追答

等价无穷小的替换

Answer 2

L=lim(x->0+) (cos√x)^(√π/x)
lnL =lim(x->0+) ln(cos√x)/ (√π/x)
=lim(x->0+) x.ln(cos√x)/ √π
=0
L = e^0 =1