Let $\alpha$ be a positive real number. Let $f: \mathbb{R} \to \mathbb{R}$ and $g: (\alpha, \infty) \to \mathbb{R}$ be the functions defined by $f(x) = \sin\left(\frac{\pi x}{12}\right)$ and $g(x) = \frac{2\log_{e}\left(\sqrt{x} - \sqrt{\alpha}\right)}{\log_{e}\left(e^{\sqrt{x}} - e^{\sqrt{\alpha}}\right)}$. Then the value of $\lim_{x \to \alpha^{+}} f(g(x))$ is ______.

$f(x) = \sin\!\left(\dfrac{\pi x}{12}\right)$

$g(x) = \dfrac{2\log_e(\sqrt{x}-\sqrt{\alpha})}{\log_e(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$

$\alpha > 0$

$\lim_{x\to\alpha^+} f(g(x)) = ?$

$1.\ t = \sqrt{x}-\sqrt{\alpha} \to 0^+$

$2.\ \text{simplify } g \text{ in terms of } t$

$3.\ \lim g = 2 \Rightarrow f(2)$

$g \text{ defined on } (\alpha,\infty) \Rightarrow \text{only } x\to\alpha^+$

$t = \sqrt{x}-\sqrt{\alpha} \to 0^+,\quad \sqrt{x} = \sqrt{\alpha}+t$

$\text{numerator} = 2\log_e t$

$e^{\sqrt{x}}-e^{\sqrt{\alpha}} = e^{\sqrt{\alpha}}(e^{t}-1)$

$g = \dfrac{2\log_e t}{\sqrt{\alpha} + \log_e(e^{t}-1)}$

$\dfrac{e^t-1}{t} \to 1 \Rightarrow \log_e(e^t-1) = \log_e t + \log_e\!\left(\dfrac{e^t-1}{t}\right)$

$\log_e\!\left(\dfrac{e^t-1}{t}\right) \to 0$

$\text{as } t\to 0^+,\ \log_e t \to -\infty$

$\sqrt{\alpha} + \log_e(e^t-1) \sim \log_e t$

$\lim_{x\to\alpha^+} g(x) = \lim_{t\to 0^+}\dfrac{2\log_e t}{\sqrt{\alpha}+\log_e(e^t-1)}$

$= \dfrac{2\log_e t}{\log_e t} = 2$

$f \text{ continuous} \Rightarrow \lim f(g(x)) = f\!\left(\lim g(x)\right) = f(2)$

$f(2) = \sin\!\left(\dfrac{2\pi}{12}\right) = \sin\dfrac{\pi}{6}$

$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$

$\boxed{\dfrac{1}{2}}$