Question

Let $\alpha$ be positive real number.
Let f:$\mathbb{R}\to\mathbb{R}$ and g:$(\alpha,\infty)\to\mathbb {R}$ be the function defined by
$\rm f(x) = \sin \bigg( \dfrac{\pi x}{12} \bigg ) \: and \: g(x) = \dfrac{2 log_{e}( \sqrt{x} - \sqrt{ \alpha } ) }{ log_{e}( {e}^{ \sqrt{x} } - {e}^{ \sqrt{ \alpha } } ) }$
Then the value of $\rm \lim_{x \to { \alpha }^ + } f(g(x))$ is​

Answer 1

First, $f(x)$ is continuous on its domain, so

$\displaystyle \lim_{x\to\alpha^+} f(g(x)) = f\left(\lim_{x\to\alpha^+} g(x)\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \sin\left(\frac\pi6 \lim_{x\to\alpha^+} \frac{\ln\left(\sqrt x - \sqrt\alpha\right)}{\ln\left(e^{\sqrt x} - e^{\sqrt\alpha}\right)}\right)$

As $x\to\alpha^+$, $\sqrt x-\sqrt\alpha\to0$ and $e^{\sqrt x}-e^{\sqrt\alpha}\to0$, so overall we have an indeterminate form ∞/∞. Apply l'Hôpital's rule and simplify.

$\displaystyle \lim_{x\to\alpha^+} f(g(x)) = \sin\left(\frac\pi6 \lim_{x\to\alpha^+} \frac{\frac1{2\sqrt x\left(\sqrt x - \sqrt\alpha\right)}}{\frac{e^{\sqrt x}}{2\sqrt x\left(e^{\sqrt x} - e^{\sqrt\alpha}\right)}}\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \sin\left(-\frac\pi6 \lim_{x\to\alpha^+} \frac{e^{-(\sqrt x-\sqrt\alpha)} - 1}{\sqrt x - \sqrt\alpha}\right)$

Substitute $y=\sqrt x-\sqrt\alpha$, so that $x\to\alpha^+\implies y\to0^+$.

$\displaystyle \lim_{x\to\alpha^+} f(g(x)) = \sin\left(-\frac\pi6 \lim_{y\to0^+} \frac{e^{-y} - 1}y\right)$

The remaining limit is the right-derivative of $e^{-y}$ at $y=0$, so

$\displaystyle \lim_{x\to\alpha^+} f(g(x)) = \sin\left(-\frac\pi6\frac{de^{-y}}{dy}\bigg|_{y\to0^+}\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \sin\left(\frac\pi6\right) = \boxed{\frac12}$