Question

If $x$ is a positive number. A sequence is defined as $x_1=\frac{1}{2}(x+\frac{3}{x}),x_2=\frac{1}{2}(x_1+\frac{3}{x_1})...x_{n+1}=\frac{1}{2}(x_n+\frac{3}{x_n}),n\ge 1,$
then $\underset{n\to \infty }{lim}{x}_n$ is equal to

• A. $3$
• B. $\sqrt{3}$
• C. $\pm \sqrt{3}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is B $\sqrt{3}$

$\underset{n\to \infty }{lim}{x}_n=\underset{n\to \infty }{lim}{x}_{n-1}=a\left(\text{say}\right)$

$\underset{n\to \infty }{lim}{x}_n=\underset{n\to \infty }{lim}\frac{1}{2}(x_{n-1}+\frac{3}{x_{n-1}})$
$\Rightarrow a=\frac{1}{2}(a+\frac{3}{a})$
$\Rightarrow a^2=3$
$\Rightarrow a=\sqrt{3}(\because x>0)$

$\therefore \underset{n\to \infty }{lim}{x}_n=\sqrt{3}$