If $f(x) =\frac{1}{3} ( \frac {5}{f(x+2)}+f(x+1))$ and $f(x) > 0$ for all $x \in \mathbb R$ then $\underset{x \to \infty}{\lim}f(x) = ?$
can anyone please give me a hint to find it?
I know how to find the limit when the limit exists. But I have no idea how to prove it's existence.
Let $\lim_{x\to\infty}f(x)$ exists. Then we can prove that:
$$\lim_{x\to\infty}f(x+1)=\lim_{x\to\infty}f(x+2)=\lim_{x\to\infty}f(x)=k$$
Then by passing limits to the both sides of your equation we get:
$$k =\frac{1}{3} ( \frac {5}{k}+k)\Leftrightarrow 3k^2=5+k^2 \Leftrightarrow k^2=\frac{5}{2}$$
Since $f(x)>0$, $k=\lim_{x\to\infty}f(x)>0$. So you get the positive solution of the above.