Algebraically or Geometrically?

Observing the Maclaurin Series of trigonometric functions, we can conclude that,

$x\to 0\implies x\to \sin x~,~x\to \tan x~,~\sin x\lt x\lt \tan x\\ \implies \frac{\sin x}{x}\to 1^-~,~\frac{\tan x}{x}\to 1^+\\ \implies \frac{a\sin x}{x}\to a^-~,~\frac{b\tan x}{x}\to b^+~\forall a,b\in \mathbb{N}$

Now, we begin to evaluate the limit using the results obtained.

$\lim_{x\to 0} \left( \left\lfloor \frac{a\sin x}{x} \right\rfloor + \left\lfloor \frac{b\tan x}{x} \right\rfloor \right)\\ =\lim_{x\to 0} \left( \left\lfloor a^- \right\rfloor + \left\lfloor b^+ \right\rfloor \right)\\ = \lim_{x\to 0} \left((a-1)+(b)\right)\\ = \boxed{a+b-1}$

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The Maclaurin series for the two trigonometric functions are:

$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots = \sum_{i=0}^\infty \frac{(-1)^i}{(2i+1)!}x^{2i+1}\\ \tan x = x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\ldots = \sum_{i=0}^\infty \frac{(-1)^i2^{2i+2}(2^{2i+2}-1)B_{2i+2}}{(2i+2)!}x^{2i+1}$

where $B_n$ is a Bernoulli number (defined only for even index).