Question

极限题求解

第四题分子搞不懂

Answer 1

构造出ln(1+x)的形式，从而利用无穷小等量代换

追答

望采纳～

Answer 2

x->1
sin(πx)
=sin(π-πx)
=sin[π(1-x)]
=π(1-x) +o(1-x)
[sin(πx)]^2 =π^2.(1-x)^2 +(o(1-x)^2)
分母
lnx
=ln(1+(x-1))
=(x-1)+o(x-1)
lim(x->1) π^2.(1-x)^2 /[(x-1)(x-1) ]

=π^2