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?

The limit is equivalent to $$\begin{align} \lim_{x\to-\infty}\left(2|x|\left(1-\frac1{4x}\right)^{1/2}+2x\right) &=\lim_{x\to-\infty}\left(-2x\left(1-\frac1{4x}\right)^{1/2}+2x\right)\\\ &=\lim_{x\to-\infty}-2x\left(\left(1-\frac1{4x}\right)^{1/2}-1\right)\\\ \end{align}$$ Then using the generalized binomial expansion we get that as $x\to0$ $$(1+x)^n=1+nx+o(x)$$ Hence our limit becomes $$\begin{align} \lim_{x\to-\infty}-2x\left(\left(1-\frac1{4x}\right)^{1/2}-1\right) &=\lim_{x\to-\infty}-2x\left(1-\frac1{8x}+o\left(\frac1x\right)-1\right)\\\ &=\lim_{x\to-\infty}-2x\left(-\frac1{8x}+o\left(\frac1x\right)\right)\\\ &=\lim_{x\to-\infty}\left(\frac14+o(1)\right)\\\ &=\frac14\\\ \end{align}$$