An Unusual Limit Problem: 10807

Abstract

Jo (1 tP)(P-l),P (1-S(X)P)1/P (1 sp)(P-1)/p' which proves the identity S(x) P + C(x)P = 1. If we differentiate the equation defining S(x), then we get S'(x) = (1S(X)P)(P l)/p = C(x)P-1, and similarly if we differentiate the equation defining C(x), then we get C'(x) = -(1-C p) (P -/p = -S(x)P-1. Thus, S'(x)C(x)-C'(x)S(x) = C(x)P + S(x)P = 1, so by the quotient rule the function T(x) = S(x)/C(x) satisfies