1个回答
2018-05-04
展开全部
设x-π/2=t
lim(x->π/2) ln[x-π/2]/tanx
=lim(t->0) lnt/tan(t+π/2)
=lim(t->0) lnt / -cot t(无穷/无穷型,用洛必达)
=lim 1/t / -(-1/(sint)^2)
=lim (sint)^2/t
=lim t
=0
lim(x->π/2) ln[x-π/2]/tanx
=lim(t->0) lnt/tan(t+π/2)
=lim(t->0) lnt / -cot t(无穷/无穷型,用洛必达)
=lim 1/t / -(-1/(sint)^2)
=lim (sint)^2/t
=lim t
=0
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