limx→π2cotx−cosx(π−2x)3 Let x=π2+t If x →π2,t→0 limt→0sint−tant−8t3 =limt→0(t−t33!+t55!+…)−(t+t33+2t515+…)−8t3 =116 We can put x=π2−t and we'll get L.H.L also same. Since L.H.L = R.H.L the limit exists and is =116