Evaluate $\lim_{x\to -\infty} (4x^2-x)^{1/2}+2x$ 
Evaluate $$\lim_{x\to -\infty} (4x^2-x)^{1/2}+2x$$

My try:-
$\lim_{x\to -\infty} (4x^2-x)^{1/2}+2x$
= $\lim_{x\to -\infty} (4x^2-x-4x^2)/ ((4x^2-x)^{1/2}-2x)$=-$\infty$
I know this is incorrect as I have seen its graph which tends to 0.25.
This means I have done a mistake. But I am unable to find it out.
 A: Note that for $x<0$
$$(4x^2-x)^{1/2}=-2x\left(1-\frac1{4x}\right)^{1/2}=-2x\left(1-\frac1{8x}+O(|x|^{-2})\right)
$$
etc.
A: Let $-\dfrac1x=h\implies h\to0^+, h>0$
$$4x^2-x=\dfrac{4-h}{h^2}\implies\sqrt{4x^2-x}=\dfrac{\sqrt{4-h}}h$$
as $\sqrt{h^2}=|h|=+h$ as $h>0$
$$\lim_{x\to-\infty}(\sqrt{4x^2-x}+2x)=\lim_{h\to0^+}\dfrac{\sqrt{4-h}-2}h=\lim_{h\to0^+}\dfrac{4-h-4}h\cdot\dfrac1{\lim_{h\to0^+}(\sqrt{4-h}+2)}=?$$
A: Let $y=-x\to \infty$
$$\lim_{x\to -\infty} \sqrt{4x^2-x}+2x=\lim_{y\to \infty} \sqrt{4y^2+y}-2y$$
and
$$\sqrt{4y^2+y}-2y=\left(\sqrt{4y^2+y}-2y\right)\frac{\sqrt{4y^2+y}+2y}{\sqrt{4y^2+y}+2y}=\frac{4y^2+y-4y^2}{\sqrt{4y^2+y}+2y}\to \frac14$$
A: $$\lim_{x\to-\infty}\sqrt{4x^2-x}+2x=\lim_{x\to-\infty}\frac{x}{2x-\sqrt{4x^2-x}}=\lim_{x\to-\infty}\frac{1}{2+\sqrt{4-1/x}}=\frac{1}{4}$$
$\textbf{Addendum:}$ The last step in the limit follows from the following 
$$2-\sqrt{4x^2-x}/x=2+\sqrt{4x^2-x}/|x|=2+\sqrt{4x^2/x^2-x/x^2}=2+\sqrt{4-1/x}$$
whenever $x<0$.
A: Your method will also work:
$$\begin{align} &\lim_\limits{x\to -\infty} \frac{-x}{\sqrt{4x^2-x}-2x}=\\
&\lim_\limits{x\to -\infty} \frac{-x}{-x\sqrt{4-\frac{1}{x}}-2x}=\\
&\lim_\limits{x\to -\infty} \frac{1}{\sqrt{4-\frac{1}{x}}+2}=\\
&\frac14.\end{align}$$
A: $(4x^2-x-4x^2)/(4x^2-x)^{1/2}=-(4x^2-x)^{1/2}/(4x-1)$, not $(4x^2-x)^{1/2}$.
Hint. Substitute $t=-x$, and multiply $\frac{(4t^2+t)^{1/2}+2t}{(4t^2+t)^{1/2}+2t}$.
