Question

Lim (sin x )^tanx as x->pi/2

Answer 1

 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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