$\lim_{x \to \infty}$

$\large \displaystyle\lim_{x \to \infty} 2^{x}\sin \dfrac{b}{2^{x}} = \lim_{x \to \infty} b \dfrac{\sin \dfrac{b}{2^{x}}}{\dfrac{b}{2^{x}}} = b \times \lim_{u \to 0} \dfrac{\sin(u)}{u} = b \times 1 = \boxed{b}$ ,

where $u = \dfrac{b}{2^{x}} \to 0$ as $x \to \infty$ .

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to \infty} 2^x \color{#3D99F6} \sin \frac b{2^x} & \small \color{#3D99F6} \text{Using Maclaurin series} \\ & = \lim_{x \to \infty} 2^x \color{#3D99F6} \left( \frac b{1! 2^x} - \frac {b^3}{3! 2^{3x}} + \frac {b^5}{5! 2^{5x}} - \frac {b^7}{7! 2^{7x}} + \cdots \right) \\ & = \lim_{x \to \infty} \left(b - \frac {b^3}{3! 2^{2x}} + \frac {b^5}{5! 2^{4x}} - \frac {b^7}{7! 2^{6x}} + \cdots \right) \\ & = \boxed{b} \end{aligned}$