Let f:R→R be a differentiable function such that f(0)=0, f(π/2)=3 and f'(0)=1. If g(x)=∫π/2x[f'(t)cosec t - cot t cosec t f(t)]dt for x∈(0,π/2], then limx→0g(x)=

g(x)=∫π/2x[f'(t) cosec t - cot t cosec t f(t)]dt = ∫π/2x f'(t)cosec t dt - ∫π/2x cot t cosec t f(t)dt
Using integration by parts for the first integral taking u(t)=cosec t and v(t)=f'(t),
g(x)=cosec t f(t)|π/2x + ∫π/2x cot t cosec t f(t)dt - ∫π/2x cot t cosec t f(t)dt
g(x)=f(π/2) cosec (π/2) - f(x) cosec (x) = 3 - f(x) cosec x = 3 - f(x)/sin x
Now, L = limx→0g(x) = 3 - limx→0 f(x)/sin x
Using L'Hospital's rule,
L = 3 - limx→0 f'(x)/cos x
L = 3 - f'(0) = 3 - 1 = 2