Question

Let $\alpha$ be a positive real number. Let $f:R\to R$ and $g:(\alpha ,\infty )\to R$ 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 $\underset{x\to {\alpha }^{+}}{\lim }f(g(x))$ is ___

Answer 1

Answer: 0.50

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