Let f   :   R → R be a differentiable function such that f 0 = 0 ,   f π 2 = 3 and f ′ 0 = 1 . If g x = ∫ x π 2   f ′ t c o s e c t - cot t   c o s e c t   f t d t , for x ∈ 0 , π 2 then lim x → 0 ⁡ g x =

Step-by-step solution

g ( x ) =   ∫ x π 2 [ f ' ( t ) c o s e c   t − f ( t ) c o s e c   t  cot   t ] d t = f ( t ) cos e c t x π 2 = f π 2 c o s e c   π 2 - f x sin ⁡ x = 3 - f x sin ⁡ x ∴           lim x → 0 ⁡ g x = 3 - lim x → 0 ⁡ f x sin ⁡ x = lim x → 0 ⁡ f ' x cos ⁡ x ;  a s   f ' 0 = 1 ⇒         lim x → 0 ⁡ g x = 3 - 1 = 2