How to prove that $x_{n+1}=g(x_n)=x_n^2+1/4\to\infty$ as $n$ increases? Let $g(x)=x^{2}+1/4$. If $g^{n}(x)=g(g(...g(x)))$ ($n$ times) how to prove for $|x|>1/2$ then $g^{n}(x)\to\infty$ as $n$ increases without bounded?  
 A: Note that $g(x)=x^2+1/4\geq x$ and equality holds iff $x=1/2$. 
This means that if $x>1/2$ then the sequence $g^{n}(x)$ is strictly increasing. 
Let $L=\sup_n g^{n}(x)=\lim_{n\to+\infty}g^{n}(x)$. 
If $L$ is finite then, $g^{n+1}(x)=g(g^n(x))$ implies that $L=g(L)$, that is $L=1/2$. Contradiction.
A: It is sufficient to notice that $g(g(x))-g(x) = x^4-\frac{x^2}{2}+\frac{1}{16}$, and the latter function is strictly positive whenever $|x|\ne \frac 12$. Therefore, the sequence $x_n$ is strictly increasing.
Suppose that this sequence is bounded, then the upper boundary is equal to the limit of this sequence, which we will denote as $x_*$. Thus, $g(x_*) = x_*$, hence $x_* = \pm \frac 12$, which is impossible (we already know that $x_*>x_2>\frac 12$).
edit
as @YvesDaoust noticed, the growth of the sequence follows from the fact that $g(x)>|x|$ whenever $|x|>\frac 12$.
A: Set $x=\frac12+a$, $a>0$. Then $g(x)=\frac12+a+a^2$. 
Playing around with some recursion steps leads to the hypothesis $$x_n=g^n(x)\ge\frac12+a+na^2,$$ which is obviously true for $x_0=\frac12+a$. Induction step: 
\begin{align}
x_{n+1}=g(x_n)
&\ge\frac12+(a+na^2)+(a+na^2)^2
\\
&\ge \frac12+a+(n+1)a^2
\end{align}
which leads to unbounded growth of the recursive sequence.

One could even find $x_n\ge\frac12+a+na^2+n(n-1)a^3$ for faster growth, but that is not necessary to prove the claim.
A: I would like to offer a direct proof. First I prove that the relation is increasing. That is  $i>n \Rightarrow g_i(x)\geq g_n(x)$. This is quite straightforward one merely needs to prove that $g(x)>x$ which comes from the fact that \begin{equation}
g(x)-x = x^2 -x+1/4=(x-1/2)^2>0
\end{equation}
Next I prove that $\forall y \exists n:g_n(x)>y$:
Consider $x>1/2\Rightarrow x=\delta+1/2$
\begin{eqnarray}
g(\delta+1/2+n\delta^2)&=&n\delta^{3}+n^2\delta^4+(n+1)\delta^2+\delta+1/2\geq(n+1)\delta^2+\delta+1/2\\
\Rightarrow g_n(1/2+\delta)&\geq&n\delta^2+\delta+1/2\\
\Rightarrow n &=& \lceil\frac{y-x}{(x-1/2)^2} \rceil+1
\end{eqnarray}
gives such an $n$.
We have thus shown that the iterations of this function are increasing and unbounded.
