Question

Let $x_n=(2^n+3^n{)}^{\frac{1}{2n}}$ for all natural numbers $n.$ Then

• A. $\underset{n\to \infty }{lim}{x}_n=\infty$
• B. $\underset{n\to \infty }{lim}{x}_n=\sqrt{3}$
• C. $\underset{n\to \infty }{lim}{x}_n=\sqrt{3}+\sqrt{2}$
• D. $\underset{n\to \infty }{lim}{x}_n=\sqrt{5}$

Answer 1

The correct option is B $\underset{n\to \infty }{lim}{x}_n=\sqrt{3}$

$\underset{n\to \infty }{lim}{x}_n=\underset{n\to \infty }{lim}(2^n+3^n{)}^{\frac{1}{2n}}=\underset{n\to \infty }{lim}{\left(3^n\right)}^{\frac{1}{2n}}\times \underset{n\to \infty }{lim}{({\left(\frac{2}{3}\right)}^n+1)}^{\frac{1}{2n}}$
$=\sqrt{3}\times 1=\sqrt{3}$