Question

$\underset{x\to \infty }{lim}{e}^{(}\frac{a^{1/x}+b^{1/x}+c^{1/x}}{3}{)}^x$, where $a,b,c$ are real and non-zero$=$

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

Answer 1

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

Let $\frac{1}{x}=y$
${lim}_{x\to \infty }exp{\left(\frac{a^{1/x}+b^{1/x}+c^{1/x}}{3}\right)}^x={lim}_{y\to 0}exp{\left(\frac{a^y+b^y+c^y}{3}\right)}^{1/y}={lim}_{y\to 0}exp\log {\left(\frac{a^y+b^y+c^y}{3}\right)}^{1/y}=exp{lim}_{y\to 0}\frac{\log \left(\frac{a^y+b^y+c^y}{3}\right)}{y}=exp{lim}_{y\to 0}\frac{1}{3}\frac{a^y+b^y+c^y-3}{y}\frac{\log (1+\frac{a^y+b^y+c^y-3}{3})}{\frac{a^y+b^y+c^y-3}{3}}$
$=exp{lim}_{y\to 0}\frac{1}{3}(\frac{a^y-1}{y}+\frac{b^y-1}{y}+\frac{c^y-1}{y})\frac{\log (1+t)}{t}$ (where $t=\frac{a^y+b^y+c^y-3}{3}$ )
$=exp\left(\frac{1}{3}(\log a+\log b+\log c)\right)={abc}^{1/3}$