Limit containing $\lfloor . \rfloor$

$\begin{aligned} L & = \lim_{x \to 0} \left(\left \lfloor 100 \cdot \frac {\sin x}x\right \rfloor + \left \lfloor 100 \cdot \frac {\tan x}x\right \rfloor \right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \left(\left \lfloor 100\cdot \frac {x-\frac {x^3}{3!}+\frac {x^5}{5!}-\cdots}x\right \rfloor + \left \lfloor 100\cdot \frac {x+\frac {x^3}3 +\frac {2x^5}{15} + \cdots}x \right \rfloor \right) \\ & = \lim_{x \to 0} \left(\left \lfloor 100 \underbrace{\left(1-\frac {x^2}{3!}+\frac {x^4}{5!}-\cdots\right)}_{\color{#D61F06}< 1} \right \rfloor + \left \lfloor 100\underbrace{\left( 1+\frac {x^2}3 +\frac {2x^4}{15} + \cdots\right)}_{\color{#D61F06}>1} \right \rfloor \right) \\ & = \left \lfloor 100 (0.999...) \right \rfloor + \left \lfloor 100 (1.000...) \right \rfloor \\ & = 99 + 100 = \boxed{199} \end{aligned}$

We combine the following facts:

• If $0 < x < \frac{\pi}{2},$ then $\sin x < x < \tan x$ and then by dividing by $x > 0$ yields $\frac{\sin x}{x} < 1 < \frac{\tan x}{x}.$ Further, since $\frac{\sin x}{x}, \frac{\tan x}{x}$ are both even functions of $x,$ we can conclude $0 < |x| < \frac{\pi}{2} \implies \frac{\sin x}{x} < 1 < \frac{\tan x}{x}$
• $\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\tan x}{x} = 1$

to conclude that $\lim_{x\to 0}\left(\left\lfloor100 \frac{\sin x}{x}\right\rfloor + \left\lfloor100 \frac{\tan x}{x}\right\rfloor\right) = \lim_{S \to 1^-}\left\lfloor100 S\right\rfloor + \lim_{T \to 1^+}\left\lfloor100 T\right\rfloor = 99 + 100 = \boxed{199}$

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Remark: In the typical development of the calculus of trigonometric functions, $\sin x < x < \tan x$ (when $0 < x < \frac{\pi}{2}$ ) is the only inequality that needs to be shown purely using the geometric definition of the trig functions. From this inequality, we can directly show $\lim_{x\to 0} \frac{\sin x}{x} = 1$ , and from this limit we can find all the derivatives of all six common trigonometric functions. The reason why I point this out is that, starting from the geometric definition of the trig functions, this is the proper way to show that $\lim_{x\to 0} \frac{\sin x}{x} = 1,$ as opposed to the common faulty argument of using L'Hopital.