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
=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
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第1个回答 2015-10-25
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本回答被网友采纳
lnL =lim(x->0+) ln(cos√x)/ (√π/x)
=lim(x->0+) x.ln(cos√x)/ √π
=0
L = e^0 =1本回答被网友采纳
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