The floor is the limit!!

$\displaystyle \lim_{x\to 0} {\left \lfloor \dfrac{\sin{x}\tan{x}}{x^2} \right \rfloor} = \lim_{x\to 0} {\left \lfloor \dfrac{\sin^2{x}}{x^2\cos{x}} \right \rfloor}$

Using Maclaurin expansion, we have,

$\displaystyle \lim_{x\to 0} {\left \lfloor \dfrac{\sin^2{x}}{x^2\cos{x}} \right \rfloor} = \lim_{x\to 0} {\left \lfloor \dfrac{\left(x-\frac{x^3}{3!} + \frac{x^5}{5!}-... \right)^2}{x^2\left(1-\frac{x^2}{2!} + \frac{x^4}{4!}-... \right)} \right \rfloor} = \left \lfloor \dfrac{x^2}{x^2\left(1 \right)} \right \rfloor = \boxed{1}$

$\lim _{ x\rightarrow 0 }{ \left\lfloor \frac { \sin { x } \times \tan { x } }{ { x }^{ 2 } } \right\rfloor } \\ =\lim _{ x\rightarrow 0 }{ \left\lfloor \frac { \sin { x } \times \frac { \sin { x } }{ \cos { x } } }{ { x }^{ 2 } } \right\rfloor } \\ =\lim _{ x\rightarrow 0 }{ \left\lfloor \frac { \sin ^{ 2 }{ x } }{ \cos { x } \times { x }^{ 2 } } \right\rfloor } \\ =\lim _{ x\rightarrow 0 }{ \left\lfloor \frac { \sin ^{ 2 }{ x } }{ \quad { x }^{ 2 } } \right\rfloor } \times \lim _{ x\rightarrow 0 } \left\lfloor \frac { 1 }{ \cos { x } } \right\rfloor \\ =\lim _{ x\rightarrow 0 }{ { \left\lfloor \frac { \sin { x } }{ x } \right\rfloor }^{ 2 } } \times \lim _{ x\rightarrow 0 } \left\lfloor \frac { 1 }{ \cos { x } } \right\rfloor \\ We\quad know\quad that\quad \lim _{ x\rightarrow 0 }{ { \frac { \sin { x } }{ x } } } =\quad 1,\quad so\quad continuing\quad on:\\ ={ \left\lfloor 1 \right\rfloor }^{ 2\quad }\times \lim _{ x\rightarrow 0 } \left\lfloor \frac { 1 }{ \cos { x } } \right\rfloor \\ =\left\lfloor \frac { 1 }{ \cos { (0) } } \right\rfloor \\ =\left\lfloor \frac { 1 }{ 1 } \right\rfloor \\ =1$