<s>[INST] 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? 
 [/INST] To avoid confusion with sign, in these cases I suggest to take $y=-x\to \infty$ then
$$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x =\lim_{y\to \infty}\sqrt{y^2-5y+3}-y $$
from here we can clearly see that the expression is an indeterminate form $\infty-\infty$ then we can proceed by
$$\sqrt{y^2-5y+3}-y=\left(\sqrt{y^2-5y+3}-y\right)\cdot \frac{\sqrt{y^2-5y+3}+y}{\sqrt{y^2-5y+3}+y}$$
Note that your first cancellation is not correct because you have considered $\sqrt{x^2+5x+3}=-x$ which is not true whereas in the second case we have cancelled out $x^2-x^2$ which is a correct step.
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