Series expansion of a function at infinity I know it is possible to expand an expandable function for a real, and for infinite by setting $x=\dfrac1y$ and then expanding for $0$. 
But my question is, how do we do if the evaluation of the new function and its derivatives is not possible ? 
I mean I find things like $\sqrt{\left(\dfrac1y\right)^2 - \dfrac1y +1}$, but I can't evaluate it at $0$ ... 
Wolfram|Alpha says it can be expanded and gives me the result which works perfectly for the rest of my problem.

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 $x\rightarrow \infty$
$$\sqrt{x^2 - x +1} = x-\frac{1}{2}+\frac{3}{8}\frac{1}{x} + \frac{3}{16}\frac{1}{x^2} + O(\frac{1}{x^3})$$
