$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.

$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*\sqrt{1-\frac{3}{x}}}$
5) $\frac{3x}{x-x(\sqrt{1-\frac{3}{x}})}$
6) $\frac{3}{1-(\sqrt{1-\frac{3}{x}})}$

Now I would just take the limit, it would result in $\frac{3}{1-1}$ which would be undefined. For some reason, the $x$ in the denominator of step 5 should turn into $-(-x)$ which in turn would be positive and therefore be $\frac{3}{1+\sqrt{1=\frac{3}{x}}}$ which would equal $\frac{3}{2}$.
I really don't get it. Apparently the $-\infty$ would mean that $\sqrt{x^2}$ = $-x$. We didn't even evaluate the limit yet.. how does that turn into $-x$, just because we know the limit is negative does not mean we evaluated it yet..., why not simplify until there is no more simplification to be done, which is what I did in my steps, which would evaluate to undefined?
Would love some help, thanks!

Using Taylor's expansion:
When x approches $-\infty$:
$x + \sqrt{x^2-3x} = x - x(1 - \frac{3}{2x}+\text{o}(x)) = x - x + \frac{3}{2}+\text{o}(1) \rightarrow \frac{3}{2}$.
So, yeah
$
\begin{align}
\lim_{x\to-\infty}x + \sqrt{x^2-3x} =\frac{3}{2}.
\end{align}
$
