Prove that $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$$
My attempt: I simplified $\sqrt{4x^2-nx}+2x$ by conjugation $$\sqrt{4x^2-nx}+2x \left(\frac{\sqrt{4x^2-nx}-2x}{\sqrt{4x^2-nx}-2x}\right) = -\frac{nx}{\sqrt{4x^2-nx}-2x} $$ I thought about substituting $x$ for $\frac{1}{y}$ to get $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \lim_{y \to 0} \frac{\sqrt{4-ny}+2}{y}$$ But this means that the limit doesn't exist! Did I go wrong somewhere in the simplification or substitution? (I noticed the $\frac{n}{4}$ part in Wolfram Alpha)
$$\lim_{x\rightarrow-\infty}-\frac{nx}{\sqrt{4x^2-nx}-2x} = \lim_{y\rightarrow \infty}\frac{ny}{\sqrt{4y^2+ny}+2y} = \lim_{y\rightarrow \infty}\frac{n}{\sqrt{4+n/y}+2} = \frac{n}{4}$$
Alternatively, $\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \lim_{y \to 0^-} \frac{-\sqrt{4-ny}+2}{y}$
Note the negative sign as $y$ is a negative quantity.