$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ 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!
 A: In such cases, to avoid confusion with signs, I often suggest, at least as a check, to take $y=-x \to \infty$ and then
$$\lim_{x\to -\infty} x+\sqrt{x^2-3x}=\lim_{y\to \infty} -y+\sqrt{y^2+3y}$$
and from here we can proceed as usual.
A: $a^2-b^2=(a-b)(a+b).$
$y:=-x$ , and consider $\lim y \rightarrow + \infty.$
$\sqrt{y^2+3y}-y= \dfrac{(y^2+3y)-y^2}{\sqrt{y^2+3y}+y}=$
$\dfrac{3y}{\sqrt{y^2+3y}+y}= \dfrac{3y}{y(\sqrt{1+3/y}+1)}$
$=\dfrac{3}{\sqrt{1+3/y}+1}.$
Take the limit.
A: The problem is indeed that, for $x<0$, $\sqrt{x^2}=-x$, so when you pull $x^2$ outside the square root it must become $-x$ and not $x$. We can assume $x<0$ because we're doing a limit for $x\to-\infty$, so restricting the function to an interval of the form $(-\infty,a)$, for any $a$, is possible and doesn't affect the limit.
You avoid most problem of this kind if you switch to positive infinity ($x=-y$) or to “positive $0$” ($x=-1/t$). With the latter method, the limit becomes
$$
\lim_{x\to-\infty}\bigl(x+\sqrt{x^2-3x}\,\bigr)=
\lim_{t\to0^+}\left(-\frac{1}{t}+\sqrt{\frac{1}{t^2}+\frac{3}{t}}\right)=
\lim_{t\to0^+}\frac{\sqrt{1+3t}-1}{t}
$$
which should be much easier (either as a derivative or using the usual technique with the conjugate). Here we can simplify $\sqrt{t^2}=t$ because we're working in a right neighborhood of $0$.
A: We are near $-\infty$, so $x<0$ and $|x|=-x$.
thus
$$x+\sqrt{x^2-3x}=x+\sqrt{x^2(1-\frac 3x)}$$
$$=x+|x|\sqrt{1-\frac 3x}$$
$$=x(1-\sqrt{1-\frac 3x})$$
$$=x\frac{\frac 3x}{1+\sqrt{1-\frac 3x}}$$
the limit is $\frac 32$.
A: 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}
$
