Question: The limit I need to solve is the following. I don't know if it's correct to treat the $+2$ and the $+1$ as insignificant, or if there's another way around it.

$$ \lim\_{x \to \infty} (\sqrt{x^2+2} - \sqrt{x^2+1})$$

Step-by-step solution will be really apreciated.

Answer: 
This has been asked many times
in many forms.

To find
$\lim \limits\_{x \to \infty} (\sqrt{x^2+a} - \sqrt{x^2+b})
$,
multiply by
$\frac{\sqrt{x^2+a} + \sqrt{x^2+b}}{\sqrt{x^2+a} + \sqrt{x^2+b}}
$
and the result is easy.
I like to do these limits by substituting $x=1/t$ (and supposing $x>0$, which is of course not restrictive); in this case, this brings the limit to the form
$$
\lim\_{t\to0^+}\frac{\sqrt{1+2t^2}-\sqrt{1+t^2}}{t}
$$
which is the derivative at $0$ of
$$
f(t)=\sqrt{1+2t^2}-\sqrt{1+t^2}
$$
Since
$$
f'(t)=\frac{2t}{\sqrt{1+t^2}}-\frac{t}{\sqrt{1+t^2}}
$$
we have $f'(0)=0$.

Or use the first order Taylor expansion of $\sqrt{1+u}=1+\frac{1}{2}u+o(u)$, so you have
$$
\lim\_{t\to0^+}\frac{(1+t^2+o(t^2))-(1+\frac{1}{2}t^2+o(t^2))}{t}=
\lim\_{t\to0^+}\left(\frac{1}{2}t+o(t)\right)=0
$$