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

Then the value of $$\lim_{x \to \alpha^+} f(g(x))$$ is ______.

We have to find $$\displaystyle\lim_{x\to\alpha^{+}} f(g(x))$$ where

$$f(x)=\sin\!\left(\frac{\pi x}{12}\right), \qquad g(x)=\frac{2\ln\!\bigl(\sqrt{x}-\sqrt{\alpha}\bigr)}{\ln\!\bigl(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\bigr)}, \qquad \alpha \gt 0.$$

Step 1 : Rewrite the limit of $$g(x)$$ as $$x\to\alpha^{+}$$.

Set $$h=\sqrt{x}-\sqrt{\alpha}, \qquad h\gt 0$$ when $$x\to\alpha^{+}.$$ Then $$\sqrt{x}=\sqrt{\alpha}+h \;\;\Longrightarrow\;\; x=(\sqrt{\alpha}+h)^{2}.$$ Consequently $$h\to0^{+}$$ is equivalent to $$x\to\alpha^{+}.$$

Step 2 : Express numerator and denominator of $$g(x)$$ in terms of $$h$$.

Numerator:

$$2\ln\!\bigl(\sqrt{x}-\sqrt{\alpha}\bigr)=2\ln h.$$

Denominator:

$$\ln\!\bigl(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\bigr) =\ln\!\Bigl(e^{\sqrt{\alpha}+h}-e^{\sqrt{\alpha}}\Bigr) =\ln\!\Bigl(e^{\sqrt{\alpha}}\bigl(e^{h}-1\bigr)\Bigr)$$

$$=\ln\!\bigl(e^{\sqrt{\alpha}}\bigr)+\ln\!\bigl(e^{h}-1\bigr) =\sqrt{\alpha}+\ln\!\bigl(e^{h}-1\bigr).$$

Step 3 : Use the first-order expansion of $$e^{h}-1$$ for small $$h$$.

For $$h\to0^{+},\;\; e^{h}-1=h+\dfrac{h^{2}}{2}+O(h^{3}).$$ Hence

$$\ln\!\bigl(e^{h}-1\bigr) =\ln h+\ln\!\Bigl(1+\dfrac{h}{2}+O(h^{2})\Bigr) =\ln h+O(h).$$

Therefore

$$\text{Denominator}= \sqrt{\alpha}+\ln h+O(h).$$

Step 4 : Evaluate the limit of $$g(x)$$.

Combine the results:

$$g(x)=\frac{2\ln h}{\ln h+\sqrt{\alpha}+O(h)} =\frac{2}{1+\dfrac{\sqrt{\alpha}+O(h)}{\ln h}}.$$

As $$h\to0^{+},\; \ln h\to-\infty\;$$ so $$\dfrac{\sqrt{\alpha}+O(h)}{\ln h}\to0.$$ Thus

$$\lim_{h\to0^{+}} g(x)=2.$$

Step 5 : Apply $$f(x)=\sin\!\left(\dfrac{\pi x}{12}\right)$$ to this limit.

$$\lim_{x\to\alpha^{+}} f\!\bigl(g(x)\bigr) =\sin\!\left(\frac{\pi}{12}\times\lim_{x\to\alpha^{+}} g(x)\right) =\sin\!\left(\frac{\pi}{12}\times2\right) =\sin\!\left(\frac{\pi}{6}\right) =\frac12.$$

Therefore the required limit equals 0.50.