Question

$\underset{x\to \infty }{lim}{\left(\frac{a^{1/x}+b^{1/x}+c^{1/x}}{3}\right)}^x=$ (where $a,b,c$ are positive real numbers)

• A. $0$
• B. $(abc{)}^{1/3}$
• C. $(abc{)}^{-1/3}$
• D. $1$

Answer 1

The correct option is B $(abc{)}^{1/3}$

$\underset{x\to \infty }{lim}{\left(\frac{a^{1/x}+b^{1/x}+c^{1/x}}{3}\right)}^x$
Put $\frac{1}{x}=t$ such that as $x\to \infty ,t\to 0$
$=e^{\underset{t\to 0}{lim}(\frac{a^t+b^t+c^t}{3}-1)\left(\frac{1}{t}\right)}$
$=e^{\underset{t\to 0}{lim}\left(\frac{a^t+b^t+c^t-3}{3}\right)\left(\frac{1}{t}\right)}$
$=e^{\underset{t\to 0}{lim}\left(\frac{(a^t-1)+(b^t-1)+(c^t-1)}{3}\right)\left(\frac{1}{t}\right)}$
$=e^{\underset{t\to 0}{lim}\left(\frac{(a^t-1)+(b^t-1)+(c^t-1)}{t}\right)\left(\frac{1}{3}\right)}$
$=e^{(\ln a+\ln b+\ln c)\left(\frac{1}{3}\right)}$
$\Rightarrow e^{\frac{\ln abc}{3}}$
$=(abc{)}^{1/3}$