Question

Consider a sequence $f\left(x\right),nϵI^{+}\cup \left\{0\right\}$ defined by $f\left(0\right)=0,f(n+1)=\sqrt{6+f\left(n\right)}$ then $\stackrel{lim}{n\to \infty }f\left(n\right)$ is

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

The correct option is C 3

as $n\to \infty ,f\left(n\right)=f(n+1)=t$
$\therefore t=\sqrt{6+t}\Rightarrow t^2-t-6=0$
$\therefore (t-3)(t+2)=0$
t = 3, t = - 2
not possible
$\therefore \stackrel{lt}{n\to \infty }f\left(n\right)=3$