Question

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

Answer 1

$\displaystyle\lim _{x \rightarrow a^{+}} \frac{2 \ln (\sqrt{x}-\sqrt{\alpha})}{\ln \left( e ^{\sqrt{x}}- e ^{\sqrt{a}}\right)}\left(\frac{0}{0} \text { form }\right)$
$\therefore$ Using Lopital rule,
$ =2 \displaystyle\lim _{x \rightarrow a^{+}} \frac{\left(\frac{1}{\sqrt{x}-\sqrt{\alpha}}\right) \cdot \frac{1}{2 \sqrt{x}}}{\left(\frac{1}{e^{\sqrt{x}}-e^{\sqrt{x}}}\right) \cdot e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}} $
$ =\frac{2}{e^{\sqrt{x}}}\displaystyle \lim _{x \rightarrow a^{+}} \frac{\left(e^{\sqrt{x}}-e^{\sqrt{x}}\right)}{(\sqrt{x}-\sqrt{\alpha})}\left(\frac{0}{0}\right)$
$ =\frac{2}{e^{\sqrt{x}}} \displaystyle\lim _{x \rightarrow a^{+}} \frac{\left(e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}-0\right)}{\left(\frac{1}{2 \sqrt{x}}-0\right)}=2$
so, $\displaystyle \lim _{x \rightarrow a^{+}} f(g(x))=\lim _{x \rightarrow a^{+}} f(2)$
$= f (2)=\sin \frac{\pi}{6}=\frac{1}{2} $
$ =0.50$