Question

Assertion :If $a_i>0\forall (i=1,2,3,...,n)$ then $\underset{x\to \infty }{lim}{\left\{\frac{{\sum }_{i=1}^n{\left(a_i\right)}^{1/x}}{n}\right\}}^{nx}=\prod _{i=1}^n{a}_i$ Reason: If $\underset{x\to a}{lim}f\left(x\right)=1$, $\underset{x\to a}{lim}g\left(x\right)=\infty$, then $\underset{x\to a}{lim}{\left\{f\left(x\right)\right\}}^{g\left(x\right)}=e^{\underset{x\to a}{lim}\{f\left(x\right)-1\}}\times g\left(x\right)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\underset{x\to \infty }{lim}{\left\{\frac{\sum {\left(a_i\right)}^{1/x}}{n}\right\}}^{nx}\forall (i=1,2,3,...,n)$
$=\underset{x\to \infty }{lim}{\left\{\frac{a_1^{1/x}+a_2^{1/x}+a_3^{1/x}+...+a_n^{1/x}}{n}\right\}}^{nx}$
$=e^{\underset{x\to \infty }{lim}\left\{\frac{a_1^{1/x}+a_2^{1/x}+a_3^{1/x}+...+a_n^{1/x}}{n}\right\}\times nx}$
$=e^{\underset{t\to 0}{lim}\left\{\frac{(a_1^t-1)+(a_2^t-1)+...+(a_n^t-1)}{n}\right\}}$ (Substitute $t=\frac{1}{x}$)
$=e^{\{\log a_1+\log a_2+\log a_3+...+\log a_n\}}$

$=e^{\log \left({\prod }_{i=1}^n{a}_i\right)}$ (i=1, 2, 3, ..., n)
$={\prod }_{i=1}^n{a}_i$