Evaluating $\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$ in two ways gives different answers 
Evaluate
$$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$$

I started by rationalising followed by dividing numerator and denominator by $x$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow-\infty}\frac{{(\sqrt{4x^2-x}+2x)}{(\sqrt{4x^2-x}-2x)}}{(\sqrt{4x^2-x}-2x)}$
=$\lim_{x\rightarrow-\infty}\frac{-x}{\sqrt{4x^2-x}-2x}$
=$\lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{4-\frac{1}{x}}-2}$
and we get $\frac{-1}{√4-2}$ i.e.-1/0 form.
The answer however is $\frac{1}{4}$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow\infty}{\sqrt{4x^2+x}-2x}$
=$\lim_{x\rightarrow\infty}\frac{{(\sqrt{4x^2+x}-2x)}{(\sqrt{4x^2+x}+2x)}}{(\sqrt{4x^2+x}+2x)}$
=$\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2+x}+2x}$
=$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{4+\frac{1}{x}}+2}$
=$\frac{1}{4}$
What is wrong in first method?

Hint note that if $x < 0$ then $-2x > 0$, but $-2 < 0$.
