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?
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x$
Apparently, the 2nd step is illegal here.
In this step, you dropped $5x+3$ to go from $\sqrt{x^2}$ to $-x$, but what makes you think you can just leave out the terms $5x+3$...? Surely you agree that $x^2+5x+3 \ne x^2$.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
How could that not be legal? The difference of two equal real numbers is $0$, of course!
The same goes for:
Also, in the 2nd step I implicitly used: $-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Addition after comment: the idea of dominating terms (in a polynomial) is fine, but you cannot choose to apply the limit "locally" and leave the other terms unchanged.
To clarify, this is fine: $$\lim_{x \to -\infty}\sqrt{x^2+5x+3}=\lim_{x \to -\infty}\sqrt{x^2}=+\infty$$ but that's not what we have here; we have: $$\lim_{x \to -\infty}\left(\sqrt{x^2+5x+3}\color{blue}{+x}\right)$$ and we cannot take the limit of the terms separately $$\lim_{x \to -\infty}\left(\sqrt{x^2+5x+3}\color{blue}{+x}\right)\color{red}{\ne}\lim_{x \to -\infty}\sqrt{x^2+5x+3}+\lim_{x \to -\infty} x$$ because both limits need to exist in order for this step to be valid.
The first step you took simply wasn’t correct.
$$\lim_{x \to -\infty}\sqrt{x^2+5x+3}+x \color{red}{\neq \lim_{x \to \infty}-x+x}$$
You shouldn’t split a limit if the individual limits don’t exist or are undefined. Going for $\infty-\infty$ is certainly incorrect as it’s an indeterminate form, so you should always avoid it.
As for your question, you’re cancelling out $x^2$ and $-x^2$. That’s simple algebra and has nothing to do with whether $x$ is approaching an infinite value or not. It always applies for all additive inverses.
You may set $x = -\frac{1}{t}$ and consider the limit for $\stackrel{t\rightarrow 0+}{\longrightarrow}$: $$\begin{eqnarray*} \sqrt{x^2+5x+3}+x & \stackrel{x = -\frac{1}{t}}{=} & \frac{\sqrt{1-5t +3t^2} - 1}{t} \\ & \stackrel{t\rightarrow 0+}{\longrightarrow} & f'(0) = -\frac{5}{2}\mbox{ for } f(t) = \sqrt{1-5t +3t^2} \end{eqnarray*}$$