The correct option is C 116
limx→π2cotx−cosx(π−2x)3
Put π−2x=t ⇒x=π2−t2
as x→π2 then t→0
=limt→0cot(π2−t2)−cos(π2−t2)t3
=limt→0tant2−sint2t3
=limt→0sint2cost2−sint2t3
=limt→0sint2(1−cost2)cost2t3
=limt→0sint2×(sin2t4)cost2×t2×2×t216×16
=limt→01cost2×sint2t2×sin2t4(t4)2×216×2
=116