Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.
But now, apparently, I could do this:
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$
which gives me the correct result.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
Also, in the 2nd step I implicitly used:
$-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Which also seems to be fine, but why? 

I will flesh out much of what would be considered a series of valid steps you may take in order to get something similar to the first step performed in the initial limit calculation. I will note that the second step that was taken is valid because we perform all of the algebra in the limit expression first, then we take a limit of what results.
The following line of reasoning shows that whenever you have $\sqrt{x^2+\beta x+\alpha}$ as a term  (added or substracted, not multiplied or divided) in a limit, it may be replaced freely by $|x+\frac{\beta}{2}|$ inside the limit (of course slight adjustments must be made for it to be shown in general).
Note that for every $b>\frac{5}{2}$, 
$$\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$
To see why this is the case, here is a series of  inequalities. First pick a $b>\frac{5}{2}$. Then pick $x$ so that $x\le\frac{3-b^2}{2b-5}$.
Then
$$(2b-5)x\le 3-b^2,$$ $$2bx+b^2\le 5x+3,$$
$$x^2+2bx+b^2\le x^2+5x+3,$$
$$\sqrt{x^2+2bx+b^2}\le\sqrt{x^2+5x+3},$$
$$\sqrt{x^2+2bx+b^2}+x\le\sqrt{x^2+5x+3}+x,$$
$$\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$
The intuition for this comes largely from working backward. So thus the claim is justified.
So now we can say that
$$\lim_{b\to \frac{5}{2}^-}\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x,$$
$$\lim_{b\to \frac{5}{2}^-}|x+b|+x=\lim_{b\to \frac{5}{2}^-}-x-b+x=\lim_{b\to \frac{5}{2}^-}-b=-\frac{5}{2}\le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$
Now since $3<\frac{25}{4}$, we have that.
$$\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x\le \lim_{x\to-\infty}\sqrt{x^2+5x+\frac{25}{4}}+x=\lim_{x\to-\infty}|x+\frac{5}{2}|+x=\lim_{x\to-\infty}-x-\frac{5}{2}+x=\lim_{x\to-\infty}-\frac{5}{2}=-\frac{5}{2}.$$
So $$\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x=-\frac{5}{2}.$$
