Playing with Limits

$\begin{aligned} A & = \lim_{n \to \infty} \lim_{x \to 0} \prod_{k=1}^n \left(1+\sin \frac x{2^k}\right)^\frac 1x \\ & = \lim_{n \to \infty} \prod_{k=1}^n \lim_{x \to 0} \left(1+\sin \frac x{2^k}\right)^\frac 1x \\ & = \lim_{n \to \infty} \prod_{k=1}^n {\color{#3D99F6} \lim_{x \to 0} \bigg(1+\frac 1{\frac 1{\sin \frac x{2^k}}} \bigg)}^{{\color{#3D99F6}\frac 1{\sin \frac x{2^k}}} \times \frac 1x \sin \frac x{2^k}} & \small \color{#3D99F6} \text{Note that }\lim_{f(x) \to \infty} \left(1+\frac 1{f(x)}\right)^{f(x)} = e \\ & = \lim_{n \to \infty} \prod_{k=1}^n {\color{#3D99F6} e}^{\lim_{x\to 0} \frac 1x \times \sin \frac x{2^k}} \\ & = \lim_{n \to \infty} \prod_{k=1}^n \exp \left(\lim_{x\to 0} \frac 1{2^k} \times \frac {\sin \frac x{2^k}}{\frac x{2^k}}\right) & \small \color{#3D99F6} \text{where }\exp (x) = e^x \\ & = \lim_{n \to \infty} \prod_{k=1}^n \exp \left(\frac 1{2^k} \right) \\ & = \lim_{n \to \infty} \exp \left(\sum_{k=1}^n \frac 1{2^k} \right) \\ & = \lim_{n \to \infty} \exp \left(1-\frac 1{2^n} \right) \\ & = e \end{aligned}$

$\begin{aligned} B & = \lim_{x \to 0} \frac {\cos^2\left(1-\cos^2 \left(1- \cos^2 \left(... \cos^2 x\right)...\right)\right)}{\sin \left(\frac {\pi(\sqrt{x+4}-2)}x\right)} \\ & = \frac {\lim_{x \to 0} \cos^2\left(1-\cos^2 \left(1- \cos^2 \left(... \cos^2 x\right)...\right)\right)}{\lim_{x \to 0} \sin \left(\frac {\pi(\sqrt{x+4}-2)}x\right)} \\ & = \frac 1{\sin \left({\color{#3D99F6}\lim_{x \to 0} \frac {\pi(\sqrt{x+4}-2)}x}\right)} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \frac 1{\sin \left({\color{#3D99F6}\lim_{x \to 0} \frac {\frac \pi{2\sqrt{x+4}}}1}\right)} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \frac 1{\sin \frac \pi 4} \\ & = \sqrt 2 \end{aligned}$

Therefore, $\lfloor A \rfloor + \lfloor B \rfloor + \lfloor A+B \rfloor = \lfloor e \rfloor + \lfloor \sqrt 2 \rfloor + \lfloor e + \sqrt 2 \rfloor = 2+1+4 = \boxed{7}$ .