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?

$\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
$-\dfrac34$.
