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.
 A: 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 $x\rightarrow \infty$
$$\sqrt{x^2 - x +1} = x-\frac{1}{2}+\frac{3}{8}\frac{1}{x} + \frac{3}{16}\frac{1}{x^2} + O(\frac{1}{x^3})$$
A: What does it mean to "expand a series at infinity?" Infinity isn't a point.  Taylor series are only defined wherever the function is k-times differentiable, and $x=1/y$ is not differentiable at zero.
The key here is differentiability; the k-th order Taylor series are only defined on an interval on which the function is k-times differentiable.  We know that differentiability implies continuity, so clearly $f$ has to be continuous wherever the Taylor series is defined.
In your example of $f(y) = \sqrt{\frac{1}{y^2} + \frac{1}{y} + 1}$, $f$ isn't continuous at 0 because it isn't defined there, and hence the Taylor series can't be expanded there.
Taylor series give us approximations of $f$ in a given neighborhood of a point $a$, so does it make sense to "approximate" $f$ at a point at which it is not continuous?
