How to find limit of $2x+ (4x^2 +3x - 2)^{1/2}$ when $x\to-\infty$ I have to find a limit $\lim _{x \to - \infty}{(2x + \sqrt{4x^2 +3x - 2})}$. I know, that the correct answer is $-3/4$, I've checked it in a graphing tool.
But could someone explain it to me, how to find it analytically, preferably without the Lhopital rule, since the one uses derivatives, and in my textbook I've not yet reached the derivatives chapter, so I'm sure the exercise should and can be solved without the Lhopital rule.
I have a hint in the textbook, that I should multiply by the conjugate. I did:
$$
\lim _{x \to - \infty}{(2x + \sqrt{4x^2 +3x - 2})} = \\
\lim _{x \to - \infty}{(\frac{4x^2 - 4x^2 -3x + 2}{2x - \sqrt{4x^2 +3x - 2}})} = \\
\lim _{x \to - \infty}{(\frac{-3x + 2}{2x - x\sqrt{4 +3/x - 2/x^2}})} = \\
\lim _{x \to - \infty}{(\frac{-3 + 2/x}{2 - \sqrt{4 +3/x - 2/x^2}})} = \\
\lim _{x \to - \infty}{(\frac{-3 - 0}{2 - \sqrt{4 - 0 + 0}})} = \\
\lim _{x \to - \infty}{(\frac{-3}{0})}
$$
That is not correct. Could anyone please, point where I'm wrong and what should I do?
 A: Your mistake, as pointed out in comments, was to take $x$ instead of $-x$ for $\sqrt{x^2}$ when $x<0$.
I like to use the binomial series:
$\lim\limits_{x\to-\infty} 2x+\sqrt{4x^2+3x-2}=\lim\limits_{x\to-\infty} 2x-2x\sqrt{1+\dfrac3{4x}-\dfrac1{2x^2}}$
$=\lim\limits_{x\to-\infty}2x-2x\left(1+\dfrac3{8x}...\right)=\dfrac{-3}4.$
A: Set $-1/x=h$ to find
$$\lim_{h\to0^+}\left(-\dfrac2h-\sqrt{\dfrac{4-3h-2h^2}{h^2}}\right)=-\lim_{h\to0^+}\dfrac{2+\sqrt{4-3h-2h^2}}h=-\dfrac{2+2}0$$ 
as for real $h,\sqrt{h^2}=|h|$ and $|h|=+h$ for $h>0$
EDIT:
$$\lim_{h\to0^+}\left(-\dfrac2h+\sqrt{\dfrac{4-3h-2h^2}{h^2}}\right)=\lim_{h\to0^+}\dfrac{\sqrt{4-3h-2h^2}-2}h$$
$$=\lim_{h\to0^+}\dfrac{4-3h-2h^2-2}h\cdot\dfrac1{\lim_{h\to0^+}(\sqrt{4-3h-2h^2}+2)}=\dfrac{-3}{\sqrt4+2}=?$$ 
A: $\begin{array}\\
ax-\sqrt{a^2x^2+bx+c}
&=(ax-\sqrt{a^2x^2+bx+c})\dfrac{ax+\sqrt{a^2x^2+bx+c}
}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{a^2x^2-(a^2x^2+bx+c)
}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{-bx-c}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{-b-c/x}{a+\sqrt{a^2+b/x+c/x^2}}\\
&=\dfrac{-b-c/x}{a+a\sqrt{1+b/(a^2x)+c/(a^2x^2)}}\\
&\to\dfrac{-b}{2a}\\
\end{array}
$
When $a=2, b=3$ this is
$-\dfrac34$.
A: The answer to this hinges on one to the question of how to extract a variable from a square root, when calculating limits, where the variable has negative values: $\lim(\sqrt{x^2}), x \in (-\infty ; 0)$
This won't qualify for a mathematical proof, but that's how I satisfied my mind.
$$
\lim_{x \to -5}(\sqrt{x^2}) \ne \lim_{x \to -5}(x) = -5
$$
So one needs to switch signs accordingly in such cases. Then
$$
\lim _{x \to - \infty}{(\frac{4x^2 - 4x^2 -3x + 2}{2x - \sqrt{4x^2 +3x - 2}})} = \\
\lim_{x \to -\infty}(\frac{-3x + 2}{2x - {\color{red} - }x \sqrt{4 + 3/x - 2/x^2}})
$$
And so on.
Thanks very much to everyone for help.
