Series expansion of a function at infinity

If you want $\sqrt{x^2 - x +1}\;$ for $x\rightarrow \infty,\;$ you have $x > 0\;$ and write $$\sqrt{x^2 - x +1} = x \sqrt{1 - \frac{1}{x} +\frac{1}{x^2}}.$$ Now substitute $y=\frac{1}{x}$ and compute the series for $y\rightarrow 0$ $$\sqrt{1 - y + y^2} = 1-\frac{1}{2}y+\frac{3}{8}y^2 + \frac{3}{16}y^3 + O(y^4)$$ Reverse the substitution, multiply by $x$ and get for $x\rightarrow \infty$ $$\sqrt{x^2 - x +1} = x-\frac{1}{2}+\frac{3}{8}\frac{1}{x} + \frac{3}{16}\frac{1}{x^2} + O(\frac{1}{x^3})$$