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

The correct answer is 0.5.

We are given two functions:

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

and

$g\left(x\right)=\frac{2{\log }_e\left(\sqrt{x}-\sqrt{\alpha }\right)}{{\log }_e\left(e^{\sqrt{x}}-e^{\sqrt{\alpha }}\right)}$

where _{$\alpha >0$}. We need to evaluate the limit:

$L=\underset{x\to {\alpha }^{+}}{\lim }f\left(g\right(x\left)\right)$

Step 1: Simplify the limit for _{$g\left(x\right)$}.
Let _{$t=\sqrt{x}-\sqrt{\alpha }$}. As _{$x\to {\alpha }^{+}$}, we have _{$t\to 0^{+}$} and _{$\sqrt{x}=\sqrt{\alpha }+t$}.
Substituting _$\sqrt{x}$ into _{$g\left(x\right)$} gives:

$g\left(x\right)=\frac{2{\log }_e\left(t\right)}{{\log }_e\left(e^{\sqrt{\alpha }+t}-e^{\sqrt{\alpha }}\right)}$

Step 2: Factor out _{$e^{\sqrt{\alpha }}$} in the denominator.
The expression inside the logarithm of the denominator can be factored as:

$e^{\sqrt{\alpha }+t}-e^{\sqrt{\alpha }}=e^{\sqrt{\alpha }}\left(e^t-1\right)$

Applying logarithmic properties to the denominator:

${\log }_e\left(e^{\sqrt{\alpha }},\left(e^t-1\right)\right)={\log }_e\left(e^{\sqrt{\alpha }}\right)+{\log }_e\left(e^t-1\right)=\sqrt{\alpha }+{\log }_e\left(e^t-1\right)$

So, _{$g\left(x\right)$} becomes:

$g\left(x\right)=\frac{2{\log }_e\left(t\right)}{\sqrt{\alpha }+{\log }_e\left(e^t-1\right)}$

Step 3: Evaluate _{$\underset{t\to 0^{+}}{\lim }g\left(x\right)$}.
Using the standard limit expansion _{$e^t-1\approx t$} as _{$t\to 0^{+}$}:

${\log }_e\left(e^t-1\right)={\log }_e\left(t·\frac{e^t-1}{t}\right)={\log }_e\left(t\right)+{\log }_e\left(\frac{e^t-1}{t}\right)$

Dividing the numerator and the denominator of _{$g\left(x\right)$} by _{${\log }_e\left(t\right)$}:

$\underset{t\to 0^{+}}{\lim }g\left(x\right)=\underset{t\to 0^{+}}{\lim }\frac{2}{\frac{\sqrt{\alpha }}{{\log }_e\left(t\right)}+1+\frac{{\log }_e\left(\frac{e^t-1}{t}\right)}{{\log }_e\left(t\right)}}$

As _{$t\to 0^{+}$}, _{${\log }_e\left(t\right)\to -\infty$} and _{$\underset{t\to 0}{\lim }\frac{e^t-1}{t}=1$}, which implies _{${\log }_e\left(1\right)=0$}.
Therefore:

$\underset{x\to {\alpha }^{+}}{\lim }g\left(x\right)=\frac{2}{0+1+0}=2$

Step 4: Compute the limit _{$\underset{x\to {\alpha }^{+}}{\lim }f\left(g\right(x\left)\right)$}.
Since the function _{$f\left(u\right)=\sin \left(\frac{\pi u}{12}\right)$} is continuous everywhere:

$\underset{x\to {\alpha }^{+}}{\lim }f\left(g\right(x\left)\right)=f\left(\underset{x\to {\alpha }^{+}}{\lim }g\left(x\right)\right)=f\left(2\right)$

Now evaluate _{$f\left(2\right)$}:

$f\left(2\right)=\sin \left(\frac{\pi ·2}{12}\right)=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}=0.5$

Thus, the value of _{$\underset{x\to {\alpha }^{+}}{\lim }f\left(g\right(x\left)\right)$} is 0.5.