\therefore \ \tan \phi = \displaystyle \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \implies \phi = \arctan \left( \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \right) = \dfrac{\pi}{2}