Floor function

$\begin{aligned} L & = \lim_{x \to 0} \left \lfloor \frac{\color{#3D99F6}{\tan^{-1} x}}{x} \right \rfloor \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \left \lfloor \frac{\color{#3D99F6}{x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... }}{x} \right \rfloor \\ & = \lim_{x \to 0} \left \lfloor \color{#3D99F6}{1 - \frac{x^2}{3} + \frac{x^4}{5} - \frac{x^6}{7} + ... } \right \rfloor \end{aligned}$

Since \(\begin{equation} \begin{split} \end{split} \color{blue}{ p(x) = 1 - \frac{x^2}{3} + \frac{x^4}{5} - \frac{x^6}{7} + ... } \end{equation} \) is even, $\displaystyle \implies \lim_{x \to 0^+} \lfloor p(x) \rfloor = \lim_{x \to 0^-} \lfloor p(x) \rfloor = L$ , the limit exists, and:

$\begin{aligned} L & = \lim_{x \to 0} \left \lfloor \color{#3D99F6}{1 - \frac{x^2}{3} + \frac{x^4}{5} - \frac{x^6}{7} + ... } \right \rfloor \quad \quad \small \color{#3D99F6}{\text{Since } p(x) < 1} \\ & = \boxed{0} \end{aligned}$