Question

Evaluate $\underset{x\to \infty }{Lim}{\left[\frac{a_1^{1/x}+a_2^{1/x}+a_3^{1/x}+......+a_n^{1/x}}{n}\right]}^{nx}$ where $a_1,a_2,a_3,..........a_n>0$

• A. $(a_1.a_2.a_3.......a_n)$
• B. $(a_1.a_3.a_5.......a_n)$ , n is odd
• C. $(a_2.a_4.a_6.......a_n)$, n is even
• D. 0

Answer 1

Answer: (A)

$(a_1.a_2.a_3.......a_n)$

$\underset{x\to \infty }{lim}{\left[\frac{a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+a_3^{\frac{1}{x}}+...a_n^{\frac{1}{x}}}{n}\right]}^{nx}=\underset{y\to 0}{lim}{\left[\frac{a_1^y+a_2^y+a_3^y+...+a_n^y}{n}\right]}^{\frac{n}{y}}$
$=exp\left[\underset{y\to 0}{lim}\frac{n}{y}[\frac{a_1^y+a_2^y+...a_n^y}{n}-1]\right]=exp\left[\underset{y\to 0}{lim}\left(\frac{a_1^y+a_2^y+...a_n^y-n}{y}\right)\right]$
$=exp\left[\underset{y\to 0}{lim}(\frac{a_1^y-1}{y}+\frac{a_2^y-1}{y}+...+\frac{a_n^y-1}{y})\right]$
$=e^{(\log a_1+\log a_2+...+\log a_n)}=e^{\log (a_1.a_2...a_n)}=a_1.a_2...a_n$