Let $A B$ be a fixed diameter of a circle of unit radius $& P$ be a variable point on this circle. Let $M$ be foot of the perpendicular drawn from the point $P$ to the tangent at point $A ,$ then $\lim _ { x \rightarrow 4 } \dfrac { \text { Area of } \Delta P A M } { ( \text { Minor arc length } P A ) ^ { 3 } }$ is equal to:

Let the angle AOP be theta. The minor arc length PA is s = theta. In the right triangle PAM, angle PAM is theta, so AM = cos(theta) * cos(theta) and PM = cos(theta) * sin(theta). The area of triangle PAM is 0.5 * AM * PM = (1/4)sin(2theta)cos(theta) = (1/2)sin(theta)cos^2(theta). Substituting theta = s and expanding sin(s) as s and cos(s) as 1, the area becomes (1/2)s(1 - s^2/2). Dividing this area by s^3 and taking the limit as s approaches 0 gives 1/4.

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