Question

$\underset{n\to \infty }{lim}{\left(\frac{\sqrt[n]{np}+\sqrt[n]{nq}}{2}\right)}^n$, $p,q>0$ equals

• A. $1$
• B. $\sqrt{pq}$
• C. $pq$
• D. $\frac{pq}{2}$

Answer 1

The correct option is B $\sqrt{pq}$

$\underset{n\to \infty }{lim}{\left(\frac{\sqrt[n]{np}+\sqrt{q}}{2}\right)}^n=\underset{n\to \infty }{lim}\left(2^{-n}{(p^{\frac{1}{n}}+q^{\frac{1}{n}})}^n\right)$

$=\underset{n\to \infty }{lim}exp\left(n\log (p^{\frac{1}{n}}+q^{\frac{1}{n}})-\log \left(2\right)n\right)$
$=exp(\underset{n\to \infty }{lim}(p^{\frac{1}{n}}+q^{\frac{1}{n}})-\left(\log 2\right)n)$

$=exp\left(\underset{n\to \infty }{lim}n\log \left(\frac{1}{2}(p^{\frac{1}{n}}+q^{\frac{1}{n}})\right)\right)$
$=exp\left(\underset{n\to \infty }{lim}\frac{\left(\log \frac{1}{2}(p^{\frac{1}{n}}+q^{\frac{1}{n}})\right)}{\frac{1}{n}}\right)$

$=exp\left(\underset{n\to \infty }{lim}\frac{p^{\frac{1}{n}}\log p+2^{\frac{1}{n}}\log 2}{p^{\frac{1}{n}}+q^{\frac{1}{n}}}\right)$
$=exp(\frac{\log p}{1+q^0}+\frac{\log q}{1+9^0})=exp(\frac{\log p}{2}+\frac{\log q}{2})=\sqrt{p}\sqrt{q}$