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29 votes
29 votes
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​

User Petem
by
2.6k points

1 Answer

10 votes
10 votes

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}

User Sarthak Gupta
by
2.6k points