A particle of mass \(m\) is constrained to move along the \(x\)-axis in a potential energy field given by \(U(x) = \dfrac{1}{2}kx^2 + \dfrac{1}{4}bx^4\), where \(k\) and \(b\) are positive constants. The particle is released from rest at \(x = A\) and oscillates symmetrically about the equilibrium position at the origin. By applying conservation of mechanical energy, the particle's maximum speed at \(x = 0\) is derived as \(v_{\max} = A\sqrt{\dfrac{k}{m} + \dfrac{bA^2}{2m}}\). Which of the following statements correctly describes the limiting behavior of \(v_{\max}\) as \(A \to 0\) and as \(A \to \infty\), along with the appropriate physical justification?