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 $x\in R$, then $\underset{x\to \infty }{lim}f\left(x\right)$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $0$

Answer 1

Answer: (B)

$2$

$f(x+2)=\frac{1}{2}\{f(x+1)+\frac{4}{f\left(x\right)}\}.....\left(i\right)$

When $x\to \infty$

$f(x+2)=f(x+1)=f\left(x\right)$

Substituting in $\left(i\right)$

$f\left(x\right)=\frac{1}{2}\{f\left(x\right)+\frac{4}{f\left(x\right)}\}2f\left(x\right)=f\left(x\right)+\frac{4}{f\left(x\right)}2f\left(x\right)=\frac{{\left(f\left(x\right)\right)}^2+4}{f\left(x\right)}2{\left(f\left(x\right)\right)}^2={\left(f\left(x\right)\right)}^2+4{\left(f\left(x\right)\right)}^2=4f\left(x\right)=\pm 2$

But $f\left(x\right)>0$ is given

So when $x\to \infty$ $f\left(x\right)=2$

Option $\left(B\right)$ is correct.