If $f(x)=\frac{1}{3} \biggl ( f(x+1)+\frac{5}{f(x+2)}\biggl)$ and $f(x)>0$, $\forall$ $x\in \mathbb R$ then $\lim_{x \rightarrow \infty} f(x)$ is? 

If $f(x)=\frac{1}{3} \biggl ( f(x+1)+\frac{5}{f(x+2)}\biggl)$ and $f(x)>0$, $\forall$ $x\in \mathbb R$ then $\lim_{x \rightarrow \infty} f(x)$ is?


Doubt:
In solution provided in book they assumed that
$\displaystyle \lim_{x \rightarrow \infty} f(x)=\displaystyle \lim_{x \rightarrow \infty}f(x+1)=\displaystyle \lim_{x \rightarrow \infty}f(x+2)=l$.
How can they all be equal?
 A: To say that $\lim_{x\to\infty} f(x) = L$ means that for all $\epsilon >0$ there exists $M > 0$ such that
$$ |f(x) - L| < \epsilon $$
whenever $ x \geq M$. Suppose that $\epsilon$ and $M$ are fixed. Then $x+1 > x \geq M$, so by the definition of the statement $\lim_{x\to\infty} f(x) = L$ it must be the case that
$$ |f(x+1) - L| < \epsilon.$$
So we obtain $\lim_{x\to\infty} f(x+1) = L$. An identical proof holds for the limit of $f(x+2)$.
As an example, consider $f(x) = \frac{1}{x}$. We have
$$\lim_{x\to\infty} \frac{1}{x} = \lim_{x\to\infty} \frac{1}{x+1} = \lim_{x\to\infty} \frac{1}{x+2} = 0.$$
The idea here is that it doesn't matter if we shift the input $x$ by some finite value, since we are allowing $x$ to grow unboundedly large.
A: A proof if we assume that $f$ has a finite limit in $+\infty$:
Let $\displaystyle \ell = \lim_{x \to +\infty} f(x)$.
We have :
$$\forall x \in \mathbb{R}, f(x) > 0$$
then $\ell \geq 0$.

*

* Suppose that $\ell = 0$ : 
We have :
$$\forall x \in \mathbb{R}, f(x) = \dfrac{1}{3} \left(f(x + 1) + \dfrac{5}{f(x + 2)}\right) > \dfrac{5}{3 f(x + 2)} \to +\infty$$
Absurd. We deduce that $\ell > 0$. 

* We have :
$$\forall x \in \mathbb{R}, f(x) = \dfrac{1}{3} \left(f(x + 1) + \dfrac{5}{f(x + 2)}\right)$$
Passing to the limit :
$$\ell = \dfrac{1}{3} \left(\ell + \dfrac{5}{\ell}\right)$$
then :
$$\ell^2 = \dfrac{5}{2}$$
We deduce that :
$$\ell = \sqrt{\dfrac{5}{2}}$$
cause $\ell \geq 0$.

