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.  
 A: We give two solutions. The first uses a standard trick that you have probably already seen. The second uses the definition of the derivative.
You  are familiar with the identity
$$(\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v,  \qquad \qquad (\ast)$$
at least in its incarnation $(a-b)(a+b)=a^2-b^2$. We will use $(\ast)$ twice. First, we multiply  the top and bottom of your expression by
$$(x +\sqrt{x^2+5x+2})(x+\sqrt{x^2+0.5x + 1}).$$
We get
$$\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\frac{(x - \sqrt{x^2+5x+2})(x +\sqrt{x^2+5x+2})(x+\sqrt{x^2+0.5x + 1})}{(x-\sqrt{x^2+0.5x + 1})(x+\sqrt{x^2+0.5x + 1})(x +\sqrt{x^2+5x+2})}.$$
The product of the first two terms on top is $-(5x+2)$, and the product of the first two terms at the bottom is $-(0.5x+1)$. It follows that 
$$\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\frac{(5x+2)(x+\sqrt{x^2+0.5x + 1})}{ (0.5x+1)(x +\sqrt{x^2+5x+2})}.\qquad\qquad(\ast\ast)$$
The behaviour of the right-hand side of $(\ast\ast)$ for large $x$ is almost obvious.  To do the calculation formally, divide top and bottom by $x^2$, as follows:
$$\frac{(5x+2)(x+\sqrt{x^2+0.5x + 1})}{ (0.5x+1)(x +\sqrt{x^2+5x+2})}=
\frac{(5+\frac{2}{x})(1+\sqrt{1+ \frac{0.5}{x} + \frac{1}{x^2}})}{ (0.5+\frac{1}{x})(1 +\sqrt{1+\frac{5}{x}+\frac{2}{x^2}})}.\qquad\qquad(\ast\ast\ast)$$
If we look at the right-hand side of $(\ast\ast\ast)$, it should be clear that as $x \to\infty$, the top approaches $10$ and the bottom approaches $1$.
Another way: Divide top and bottom by $x$. We get 
$$\frac{1 - \sqrt{  1+\frac{5}{x}+\frac{2}{x^2}}} {1-\sqrt{  1+\frac{0.5}{x} + \frac{1}{x^2} }}.$$
Make the substitution $t=\frac{1}{x}$, and change the sign of top and bottom. We are now interested in the limit as $t$ approaches $0$ from the right of 
$$\frac{\sqrt{1+5t+2t^2}-1} {\sqrt{1+0.5t + t^2 }-1}.$$
Looks nice than before! We could now use a trick much like in the first solution. But we will use another idea. Multiply top and bottom by $t-0$. We get 
$$\frac{\sqrt{  1+5t+2t^2}-1}{t-0}\frac{t-0}{\sqrt{  1+0.5t + t^2 }-1}.$$
What is 
$$\lim_{t\to 0}\frac{\sqrt{  1+5t+2t^2}-1}{t-0}?$$
If we look back on the definition of the derivative, we will recognize 
this limit as the derivative of $f(t)=\sqrt{1+5t+2t^2}$ at $t=0$. Calculate $f'(0)$. It is $5/2$.  The limit of the second term is $1/g'(0)$, where $g(t)=1+0.5t +t^2$. Calculate. We get $1/(0.5/2)$. The product of the two limits is $10$.
Comment: The technique (in another answer) that uses the Taylor expansion is more sophisticated, but no harder.  In the long run it is much more useful.
A: Divide through by $x, \lim \limits_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}}$ then use the expansion around $x=\infty \lim  \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}} \approx \frac{1-[1+5/(2x)]}{1-[1+1/(4x)]}=10$ where the $\approx$ means here that the first orders are the same.
A: 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 $u \to \infty$.
