Not kidding this time

$\\ \lim_{x \to 0} \left \lfloor \frac{-2x}{\tan x} \right \rfloor = \\\\\\ \lim_{x \to 0} \left \lfloor \frac{- 2x}{\frac{\sin x}{\cos x}} \right \rfloor = \\\\\\ \lim_{x \to 0} \left \lfloor - 2 \cdot \cos x \cdot \frac{x}{\sin x} \right \rfloor = \\\\\\ \lim_{x \to 0} \left \lfloor - 2 \cdot \cos x \cdot \left (\frac{\sin x}{x} \right )^{- 1} \right \rfloor =$

Do limite fundamental, $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . Temos que:

$\\ \lim_{x \to 0} \left \lfloor - 2 \cdot \cos x \cdot \left (\frac{\sin x}{x} \right )^{- 1} \right \rfloor = \\\\\\ - 2 \cdot \cos 0^o \cdot (1)^{- 1} = \\\\ - 2 \cdot 1 \cdot 1 = \\\\ \boxed{- 2}$

Let $L = \dfrac{-2x}{\tan x}$ , then:

$\begin{aligned} \lim_{x \to 0} \lfloor L \rfloor & = \lim_{x \to 0} \left \lfloor \frac{-2x}{\color{#3D99F6}{\tan x}} \right \rfloor \\ & = \lim_{x \to 0} \left \lfloor \frac{-2x}{\color{#3D99F6}{x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+...}} \right \rfloor \quad \quad \small \color{#3D99F6}{\text{Maclaurin series}} \\ & = \lim_{x \to 0} \left \lfloor \frac{-2}{1+\frac{1}{3}x^2+\frac{2}{15}x^4+\frac{17}{315}x^6+...} \right \rfloor \end{aligned}$

We note that as $x \to 0$ , $L \to -2^+ > - 2$ , therefore $\displaystyle \lim_{x \to 0} \lfloor L \rfloor = \boxed{- 2}$ .

As x tends to 0 ,x/tan x=1 so the limit is [-2] where [] denotes floor function & [-2]=-2