Given that $f(x)=x^2+\frac{1}{4}$, there exists the iterated sequence ${f^{\circ n}(x)}_{n=1}^\infty$ (where $f^{\circ n}(x)$ is defined as $\underbrace{f(f(f...(x)...))}_{n\text{ times}}$), which is increasing on the interval $(1/2, \infty)$. I need to prove that this sequence is not bounded on this interval, without actually knowing the sequence explicitly.
Can you please give me a hint as to how such a proof would work? I still have no idea, besides the fact that it is possible to do the proof by just using the function $f(x)$ above. I've tried to do it using the $\delta$-$\varepsilon$ definition, but apparently couldn't find a direction to follow.
If it is bounded, since is increasing, then it is convergent. Let $\ell$ it's limit.
By definition, $x_{n+1}=f(x_n)=x_n^2+\frac{1}{4}$, so $\ell$ satisfies the relation $\ell=\ell^2+\frac{1}{4}$. Thus $\ell=\frac{1}{2}$.
Now, since $\{x_n\}$ is increasing, then $\ell=\frac{1}{2}<x_0\le x_n\le\ell$ (in every convergert increasing sequence, the limit is greater than the elements of the sequence). This is a contradiction. Thus the sequence is unbounded.