Divergence of a recursive sequence If $(x_n)$ isthe sequence defined by $x_1=\frac{1}{2}$ and $x_{n+1}=\sqrt{x_n^2 +x_n +1}$, show that $\lim x_n = \infty$
Ive tried a couple of things but none of them helped. Ive tried to suppose, by contradiction, that the sequence is bounded, find a lower sequence that goes to infinite and the definition. 
Thanks in advance! 

The sequence is increasing. If it were bounded above, it would have a limit $L$. Then
$$L=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty} \sqrt{x_n^2+x_n+1}=\sqrt{L^2+L+1}$$ would give $L=\sqrt{L^2+L+1}$, which is impossible, since $L\gt 0$.  
