<s>[INST] Value of $\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$ The value of the $$\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$$ is?
The answer given is $1/4$.
I rationalized and got $$\lim\limits_{x\to-\infty} \frac {-x}{|x|[(4- \frac {1}{x})^{1/2}-2]}$$ how to proceed further?
 [/INST] \begin{align*}
(4x^2-x)^{1/2}+2x
&=\sqrt{4x^2-x}+2x
\\&=\frac{(\sqrt{4x^2-x}+2x)(\sqrt{4x^2-x}-2x)}{\sqrt{4x^2-x}-2x}
\\&=\frac{(4x^2-x)-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{(2x)^2(1-\frac{1}{4x})}-2x}
\\&=\frac{-x}{2|x|\sqrt{1-\frac{1}{4x}}-2x}
\\&\qquad [x<0]
\\&=\frac{-x}{-2x\sqrt{1-\frac{1}{4x}}-2x}
\\&=\frac{1}{2\sqrt{1-\frac{1}{4x}}+2}
\\&\to\frac{1}{2\sqrt{1+0}+2}
\\&=\frac{1}{4}.
\end{align*}
(as $x\to-\infty$)
</s>