lim (x → π/2) (sinx)^(tanx)
= lim (x → π/2) e^[(tanx) ln (sinx)]
= e^ [lim (x → π/2) (tanx) ln (sinx)] ... (1)
lim (x → π/2) (tanx) ln (sinx)
= lim (x → π/2) [ln(sinx) / cotx]
Using L'Hospital'stheorem,
= lim (x → π/2) [- cotx / cosec^2 x]
= 0
Plugging in ( 1 ),
required limit = e^0 = 1
=>Answer is 1.
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