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! 

$$x_{n+1}=\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}>x_n+\frac{1}{2}$$
This implies that $\frac{1}{2}n\leq x_n$ so $x_n$ diverges to $\infty$
