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?
 A: The problem happens when you go from
$$\lim_{x\rightarrow-\infty}\frac{-x}{\sqrt{4x^2-x}-2x}$$
to
$$\lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{4-\frac{1}{x}}-2}$$
The first line is actually not identical to the second line, and you can verify yourself by plugging in some value of $x$. For example, if $x=-20$, then
$$\frac{-x}{\sqrt{4x^2-x}-2x} = \frac{-(-20)}{\sqrt{4\cdot(-20)^2-(-20)}-2\cdot(-20)} = \frac{20}{\sqrt{1620}+40}\approx\frac{20}{40+40}=\frac14$$
while
$$\frac{-1}{\sqrt{4-\frac{1}{x}}-2} = \frac{-1}{\sqrt{4-\frac{1}{-20}}-2}=\frac{-1}{\sqrt{4+\frac{1}{20}}-2}\approx\frac{-1}{2+\epsilon-2}=\frac{-1}{\epsilon}$$ which is very negative, because $\epsilon$ is small.

So, you need to be very careful when going from one expression to the other. In particular, what you probably did was this:
$$\begin{align}
\frac{-x}{\sqrt{4x^2-x}-2x}&=\frac{\frac{-x}{x}}{\frac{\sqrt{4x^2-x}-2x}{x}}\\
&=\frac{-1}{\frac{\sqrt{4x^2-x}}{x}-\frac{2x}{x}}\\
&=\frac{-1}{\frac{\sqrt{4x^2-x}}{\sqrt{x^2}}-2}\\
&=\frac{-1}{\sqrt{\frac{4x^2-x}{x^2}}-2}\\
&=\frac{-1}{\sqrt{4-\frac1x}-2}
\end{align}$$
And you can see (again, just plug in $x=20$ for a sanity check) that only the first two lines of the above expression are correct.
The mistake comes when you replace $x$ with $\sqrt{x^2}$, which is an equality that does not hold for $x<0$. Instead, you must replace $x$ with $-\sqrt{x^2}$ to get
$$\frac{-1}{\frac{\sqrt{4x^2-x}}{-\sqrt{x^2}}-2} = \frac{-1}{-\sqrt{4-\frac1x}}-2$$
and then you get the correct result.
A: Hint note that if $x < 0$ then $-2x > 0$, but $-2 < 0$.
