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( (\pi x)/(12) \bigg ) \: and \: g(x) = \frac{2 log_(e)( √(x) - √( \alpha ) ) }{ log_(e)( {e}^( √(x) ) - {e}^( √( \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^+) (\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)}}{(e^(\sqrt x))/(2\sqrt x\left(e^(\sqrt x) - e^(\sqrt\alpha)\right))}\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \sin\left(-\frac\pi6 \lim_(x\to\alpha^+) (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(de^(-y))/(dy)\bigg|_(y\to0^+)\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \sin\left(\frac\pi6\right) = \boxed{\frac12}`