$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$ $$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$$
What I did
I multiplied by the denominator's conjugate and got the following $$2x+\sqrt{4x^2-2x}$$
My question is,  what would I now do to evaluate the limit? The positive infinity I would do as follows:
$$\lim_{x\rightarrow\infty}2x+\sqrt{4x^2-2x} = 2\cdot\infty + \sqrt{4\cdot(\infty)^2 - 2\cdot\infty} = \infty???$$

When $x \to +\infty$
$$2x+\sqrt{4x^2-2x} = 2x\left(1 +\sqrt{1- \dfrac{1}{2x}}\right) \sim 4x\to +\infty$$
When $x \to -\infty$
$$2x+\sqrt{4x^2-2x} = 2x\left(1 -\sqrt{1- \dfrac{1}{2x}}\right) = \dfrac{1}{1 +\sqrt{1- \dfrac{1}{2x}}} \to \dfrac{1}{2}$$
