Let α be a positive real number. Let f: ℝ → ℝ and g: (α, ∞) → ℝ be...

$\lim_{x \to {\alpha}^{+}}g\left(x\right) = \lim_{x \to {\alpha}^{+}}\dfrac{2\log_{e}\left(\sqrt{x} - \sqrt{\alpha}\right)}{\log_{e}\left({e}^{\sqrt{x}} - {e}^{\sqrt{\alpha}}\right)} = \lim_{x \to {\alpha}^{+}}\dfrac{\dfrac{2}{\sqrt{x} - \sqrt{\alpha}}\left(\dfrac{1}{2\sqrt{x}}\right)}{\dfrac{1}{{e}^{\sqrt{x}} - {e}^{\sqrt{\alpha}}}\left(\dfrac{1}{2\sqrt{x}}{e}^{\sqrt{x}}\right)}$

$= \lim_{x \to {\alpha}^{+}}\dfrac{{e}^{\sqrt{x}} - {e}^{\sqrt{\alpha}}}{\sqrt{x} - \sqrt{\alpha}}\dfrac{1}{{e}^{\sqrt{x}}} \cdot 2$

$= \dfrac{2}{{e}^{\sqrt{\alpha}}}\lim_{x \to {\alpha}^{+}}\dfrac{{e}^{\sqrt{x}} \cdot \dfrac{1}{2\sqrt{x}}}{\dfrac{1}{2\sqrt{x}}} = 2$

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

$= f\left(2\right) = sin\dfrac{\pi}{6} = \dfrac{1}{2} = 0 . 50$