Evaluate the limit of  $\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$ $$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$
Can't find a means to resolve. The answer is $10$ by graphing.  

Completing the squares:  
$x^2+5x+2 = (x + 2.5)^2 + O(1)$  
$x^2+0.5x+1 = (x + 0.25)^2 + O(1)$
So:
$\sqrt{x^2+5x+2} = x + 2.5 + o(1)$  
$\sqrt{x^2+0.5x+1} = x + 0.25 + o(1)$
Now the answer is immediate.
Updated to address didier's concerns:  
We show using elementary methods that $\sqrt{u^2+O(1)}=u+o(1)$, which is the same as saying that for any constant $C$, $\sqrt{u^2+C} \to u$ as $u \to \infty$.  
Lemma: If $|\epsilon| < 1$, then $1 - |\epsilon| \le \sqrt{1 + \epsilon} \le 1 + |\epsilon|$.  
Proof: If $\epsilon \ge 0$, then 
$$1 - \epsilon \le 1 \le \sqrt{1 + \epsilon} \le 1 + \epsilon$$
because $\sqrt a \le a$ if $a \ge 1$.  
If $\epsilon < 0$, put $\theta = -\epsilon$. Then
$$1 - |\epsilon| = 1 - \theta < \sqrt{1 - \theta} = \sqrt{1 + \epsilon}$$
because $a < \sqrt a$ if $0 < a < 1$. And
$$\sqrt{1+\epsilon} = \sqrt{1-\theta} < 1 < 1 + |\epsilon|$$
Corollary: If $|\epsilon| < 1$, then $|\sqrt{1+\epsilon} - 1| \le |\epsilon|$.  
Now we just put $\epsilon = C/u^2$ in the corollary to get (for $u > \sqrt{|C|}$):
$$|\sqrt{u^2 + C} - u| = u\Big|\sqrt{1 + \frac{C}{u^2}} - 1\Big| \le u\frac{|C|}{u^2} = \frac{|C|}{u}$$
which tends to $0$ as $u \to \infty$.
