Question

If $f\left(x\right)=\frac{1}{3}\left\{f\right(x+1)+\frac{5}{f(x+2)}\}$ and $f\left(x\right)>0$ for all $x$ $\in$ R,

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

• A. $\sqrt{\frac{2}{5}}$
• B. $\sqrt{\frac{5}{2}}$
• C. $\infty$
• D. None of the above

Answer 1

Answer: (B)

$\sqrt{\frac{5}{2}}$

Let $\underset{x\to \infty }{lim}f\left(x\right)=l$, then
$\underset{x\to \infty }{lim}f(x+1)=\underset{x\to \infty }{lim}f(x+2)=l$.
Now,
$f\left(x\right)=\frac{1}{3}\left\{f\right(x+1)+\frac{5}{f(x+2)}\}$
$\Rightarrow {lim}_{x\to \infty }f\left(x\right)=\frac{1}{3}\left\{{lim}_{x\to \infty }f\right(x+1)+\frac{5}{f(x+2)}\}$
$\Rightarrow l=\frac{1}{3}(l+\frac{5}{l})$
$\Rightarrow 3l^2=l^2+5\Rightarrow 2l^2=5\Rightarrow l=\sqrt{\frac{5}{2}}$
Hence, option 'B' is correct.