$\displaystyle\lim_{x\to-\infty}2x+\sqrt{4x^2+x}$ How do I calculate
$$\lim_{x\to-\infty}2x+\sqrt{4x^2+x}?$$
I've gotten "close" many times but I just can't figure it out, I always get stuck with zero division or something similar.

$$\begin{align*} \lim_{x \rightarrow -\infty} 2x + \sqrt{4x^2 + x} & = \lim_{x \rightarrow \infty} \sqrt{4x^2 - x} - 2x = \\ & = \lim_{x \rightarrow \infty} \frac{(\sqrt{4x^2 - x} - 2x)(\sqrt{4x^2 - x} + 2x)}{\sqrt{4x^2 - x} + 2x} = \\ & = \lim_{x \rightarrow \infty} -\frac{x}{\sqrt{4x^2 - x} + 2x} = \boxed{-\frac{1}{4}} \end{align*}$$
