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?

Question

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?

Answer 1

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.

Answer 2

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 $\ell \geq 0$.

1. 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 $\ell > 0$.
2. 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 $\ell \geq 0$.