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Question Number 21223 by moadakennaf last updated on 16/Sep/17

lim_(x→π/2) (π−2x)tan (x)

limxπ/2(π2x)tan(x)

Answered by dioph last updated on 16/Sep/17

= lim_(x→π/2)  ((π−2x)/(cot x)) =  = lim_(x→π/2)  ((−2)/(−cosec^2  x)) =  = lim_(x→π/2)  2 sin^2  x =  = 2

=limxπ/2π2xcotx==limxπ/22cosec2x==limxπ/22sin2x==2

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