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

$\lim\limits_{x→α^+}f(g(x))=f(\lim\limits_{x→α^+}g(x))$
Now, $\lim\limits_{x→α^+}g(x)=\lim\limits_{x→α^+}\frac{2ln(\sqrt{x}-\sqrt{\alpha})}{ln(e^{\sqrt{x}}-e^{\sqrt{α}})}$
Now, By applying D'L Hospital
$\lim\limits_{x→α^+}+\frac{2.\frac{1}{\sqrt{x-\sqrt{\alpha}}}.\frac{1}{2\sqrt{x}}}{\frac{1}{e^{\sqrt{x}}=e^{\sqrt{\alpha}}}.e^{\sqrt{x}}.\frac{1}{2\sqrt{x}}}$
$\lim\limits_{x→α^+}\frac{2(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}{e^{\sqrt{x}}(\sqrt{x}-\sqrt{\alpha})}$
$\lim\limits_{x→α^+}\frac{2e^{\sqrt{\alpha}}(e^{\sqrt{x}-\sqrt{\alpha}}-1)}{e^{\sqrt{x}}(\sqrt{x}-\sqrt{\alpha})}=2$
Now, f(x) = $\sin\frac{\pi x}{12}$ given
Hence f(2) = $\sin\frac{\pi(2)}{12}$
$=\sin\frac{\pi}{6}$
$=\frac{1}{2}=0.5$
So, the correct answer is 0.5