Question

求此题做法，。？。

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Answer 1

L =lim(x->π/2) (sinx)^tanx

lnL
=lim(x->π/2) ln(sinx)/cotx (0/0)
=lim(x->π/2) [cosx/sinx]/-(cscx)^2
=lim(x->π/2) -[cosx.sinx]
=0
=>
L =e^0 =1