<s>[INST] 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.  
 [/INST] 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.
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