Question

The value of $\underset{x\to \infty }{\lim }{\left(\frac{a_1^{1/x}+a_2^{1/x}+...........+a_n^{1/x}}{n}\right)}^{nx}{a}_i>0,i=1,2,......n$, is

• A. $a_1+a_2+...........+a_n$
• B. $e^{a_1+a_2+...a_n}$
• C. $\frac{a_1+a_2+...+a_n}{n}$
• D. $a_1{a}_2{a}_3.......a_n$

Answer 1

Answer: (D)

Putting $x=\frac{1}{y}$, we get
$L=limit=\underset{y\to 0}{\lim }{\left(\frac{a_1^y+a_2^y+.....+a_n^y}{n}\right)}^{n/y}\left(\because x\to \infty y\to 0\right)$
$\therefore lo{g}_{e}L=\underset{y\to 0}{\lim }\frac{n}{y}.lo{g}_e\frac{1}{n}\left(a_1^y+a_2^y+.....+a_n^y\right)\left(\frac{0}{0}\right)$
$=n\underset{y\to 0}{\lim }\frac{\frac{\left(a_1^{y}log{a}_1+a_2^{y}log{a}_2+.....+a_n^{y}log{a}_n\right)}{a_1^y+a_2^y+.....+a_n^y}}{1}$
$=n.\frac{log\left(a_1{a}_2....a_n\right)}{n}$
$\therefore logL=log\left(a_1.a_2.....a_n\right)\Rightarrow L=a_1.a_2.a_3......a_n$