proof of limit for $x \to +\infty = +\infty$ My exercise book proves the limit: $lim_{x \to +\infty} \frac{x^2 - 2}{2x + 1} = +\infty$ in this way: $\lor K > 0, \exists \bar{x} = \bar{x}_k$ : if $x > \bar{x} \Rightarrow f(x) > K$
Then, the author solved the inequality $\frac{x^2 - 2}{2x + 1} > K$
Which is valid for $x > \sqrt{K^2 + K + 2} \lor x < K - \sqrt{K^2 + K + 2} \lor x > -\frac{1}{2}$
But the author writes the solution this way:
$K - \sqrt{K^2 + K + 2} < x < -\frac{1}{2} \lor x > \sqrt{K^2 + K + 2}$
How can he say that $K - \sqrt{K^2 + K + 2} < -\frac{1}{2}$
What assumptions do he make?
Also, the exercise ends with $\bar{x} = K + \sqrt{K^2 + K + 2}$
why $K + \sqrt{K^2 + K + 2}$ and not $K - \sqrt{K^2 + K + 2}$?
Thank you.
 A: First we find a simpler argument. Note that  $x^2-2=\frac{x^2}{2}+\frac{x^2}{2}-2\ge \frac{x^2}{2}$ if $x\ge 2$. 
Note also that if $x\ge 1$ then $2x+1\le 3x$. It follows that if $x\ge 2$, then
$$\frac{x^2-2}{2x+1} \ge \frac{x^2/2}{3x}=\frac{x}{6}.$$
Thus if $x\gt \max(2, 6K)$, then $\frac{x^2-2}{2x+1}\gt K$.
The point is that we need not try to find the cheapest $\bar{x}$ such that beyond $\bar{x}$ our function is $\gt K$. All we need to do is to show that there exists such a $\bar{x}$.
Now to the actual question. Suppose that we want to solve the inequality $\frac{x^2-2}{2x+1}\lt K$. Note that in this context that is not a good thing to do.  
There are various approaches. The one that seems to have been taken is to separate into cases, $x\gt -1/2$ and $x\lt -1/2$.
The case $x\lt -1/2$ is obviously irrelevant. But if we insist on solving in that case, we can rewrite the inequality as $x^2-2\gt K(2x+1)$ (we are multiplying both sides by the negative quantity $2x+1$).
In the case $x\gt -1/2$, we can rewrite our inequality as $x^2-2\lt K(2x+1)$. The roots of the equation $x^2-2Kx-K-2=0$ are $x=K\pm\sqrt{K^2+K+2}$.  
You ask why $K-\sqrt{K^2+K+2} \lt -1/2$. That question does not need to be answered to solve our problem. But to answer it, note that $K^2+K+2=(K+1/2)^2 +7/4$, so $\sqrt{K^2+K+2} \gt K+1/2$. Thus $K-\sqrt{K^2+K+2}\lt K-(K+1/2)=-1/2$.
But we are interested in finding a place such that beyond it our inequality holds. That place is the larger root $K+\sqrt{K^2+K+2}$.  
