Question

If $f(x+2)=\frac{1}{2}\{f(x+1)+\frac{4}{f\left(x\right)}\}$ and $f\left(x\right)>0$ for all $xϵR$ then $\underset{x\to \infty }{lim}f\left(x\right)$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $0$

Answer 1

The correct option is B $2$

Let $l=\underset{x\to \infty }{lim}f\left(x\right)$
Now as, $x\to \infty ,x+1,x+2\to \infty =>f(x+1),f(x+2)\to l$
Thus given equation becomes, $l=\frac{1}{2}(l+4/l)$
$\Rightarrow 2l^2=l^2+4$
$\Rightarrow l^2=4$
$\Rightarrow l=2$,as $f\left(x\right)>0$