limits as x goes to negative infinity

Question

What is the limit of this:

$\lim_{x \to -\infty} (4x^2 + 3x)^\frac 12 + 2x$

I tried to use conjugates and then divide by the highest degree of $x$ but I got $\frac 30$ which is not the right answer of $-\frac 34$. Is the process that I took correct, can I use algebra or do you have to think logically?

Answer 1

$$\lim_{x\rightarrow-\infty}\left(\sqrt{4x^2+3x}+2x\right)=\lim_{x\rightarrow-\infty}\frac{4x^2+3x-4x^2}{\sqrt{4x^2+3x}-2x}=$$ $$=\lim_{x\rightarrow-\infty}\frac{3}{-\sqrt{4+\frac{3}{x}}-2}=-\frac{3}{4}.$$

Answer 2

You have $$\lim_{x \to -\infty} (4x^2 + 3x)^\frac 12 + 2x =\lim_{x \to \infty} (4x^2 - 3x)^\frac 12 - 2x. $$ Now \begin{align} (4x^2 - 3x)^\frac 12 - 2x&=\frac{4x^2-3x-4x^2}{(4x^2-3x)^{1/2}+2x} =\frac{-3x}{(4x^2-3x)^{1/2}+2x}\\ \ \\ &=\frac{-3}{(4-3/x)^{1/2}+2}\xrightarrow[x\to\infty]{}-\frac34. \end{align}