Limit when the function is given indirectly

Question

Suppose we have a function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined recursively by $$f(x)=\frac{1}{3}\left[f(x+1)+\frac{5}{f(x+2)}\right] $$

Assuming $~f(x) >0$ for all $x\in\mathbb{R}$, compute $\lim_{x\rightarrow\infty}f(x)$

Answer 1

Assuming $\lim_{x\rightarrow\infty}f(x)$ exists and is finite, we have $$ \lim_{x\rightarrow\infty} f(x)=\frac{1}{3}\left[\lim_{x\rightarrow\infty} f(x+1)+\frac{5}{\lim_{x\rightarrow\infty}f(x+2) }\right]. $$ Since $\lim_{x\rightarrow\infty}f(x+c) =\lim_{x\rightarrow\infty}f(x)$ for all $c\in\mathbb{R}$ (since $c$ is negligible with respect to $\infty$), we have $$ \lim_{x\rightarrow\infty} f(x)=\frac{1}{3}\left[\lim_{x\rightarrow\infty} f(x)+\frac{5}{\lim_{x\rightarrow\infty}f(x) }\right]. $$

Let $a:=\lim_{x\rightarrow\infty} f(x)$, the above equation yields $$ 3a=a+\frac{5}{a}, $$ which gives $a=\pm\sqrt{\frac{5}{2}}$ as a result. We know that $f(x)>0$ for all $x$, so we conclude $a=+\sqrt{\frac{5}{2}}$ which is what we expected.

Answer 2

Let $L$ denote this limit, assuming it exists. Then taking the limit of both sides of your equation, $$L = \frac{1}{3}(L + \frac{5}{L}).$$ Using this data and the fact that $f(x) > 0$ for every $x$, we can use algebra to solve for $L$.