44671970
幼苗
共回答了21个问题采纳率：100% 举报
原式=lim(x→1) (√x-1)(√x+1)[x^(1/3)-1][x^(2/3)+x^(1/3)+1]/{(√x+1)[x^(2/3)+x^(1/3)+1]*[1-cos(π-πx)]}(分子有理化)
=lim(x→1) (x-1)^2/{(√x+1)[x^(2/3)+x^(1/3)+1]*(π-πx)^2/2}(等价无穷小代换)
=lim(x→1) 2/{(√x+1)[x^(2/3)+x^(1/3)+1]*π^2}
=1/(3π^2)
1年前
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