How to deduce the radius of convergence of a rational function on the reals?

Question

What arguments are available to deduce that the radius of convergence of the Taylor series of $$f(x)=\frac{x^3-1}{x-2}$$ centered at $x=0$ is 2, where $f$ is considered as a real function of a real variable?

Note: I am aware that the answer is the distance to the closest singularity, but I do not want an invocation of that fact. What arguments could you use to get to the result if you had forgotten this fact?

Observe that the series for $x^3-1$ (namely itself) has radius $\infty$, while the series for $(x-2)^{-1}$ has radius 2. Since the radius of convergence of the product of two power series is at least the minimum of the radii of the factors, and since Taylor series are multiplicative, we conclude that the radius of convergence of the Taylor series of $f$ at $x=0$ is at least 2.

The harder part is showing it can't be more than 2. It is obvious that $f$ can't equal its Taylor series past 2 because of the discontinuity, but this alone does not imply that the Taylor series for $f$ can't converge past 2. The result follows, though, as soon as we show $f$ equals its Taylor series out to 2.

So far as I can see, there are two possibilities:

1. argue by direct calculation

This approach ignores the observations above; we simply compute. The Taylor series of the product $(x^3-1)(x-2)^{-1}$ is the product of the Taylor series of the factors,

$$(x^3-1)\cdot\left(-\frac{1}{2}\right)\left(1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+O(x^4)\right)=\frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}+\sum_{k=3}^\infty\left(2^{-k-1}-2^{-k+2}\right)x^k$$

The ratio test shows this converges when $|x|<2$.

2. argue using Cauchy's theorem on the expansion of a function in a power series in its domain of analyticity (theorem 16.7 in Markushevich)

Since the corresponding complex function is differentiable in the disk $|z|<2$, Cauchy's theorem says it equals its Taylor series on that disk; hence the same is true for the real function $f$, which must therefore equal its Taylor series out to $x=2$. Since $f$ has an infinite discontinuity at $2$ but Taylor series are continuous, the series can't converge past 2.

Are there other approaches that don't amount to these two in disguise? I'm curious whether I'm missing something.

Answer 1

It's not a "heuristic". The function $f(z) = \dfrac{z^3-1}{z-2}$ is analytic on the complex plane except for the singularity at $z=2$, which is a pole (note that $|f(z)| \to \infty$ as $z \to 2$). Therefore its Taylor series about $z=a$ has radius of convergence $|a-2|$.

Answer 2

Boring approach: $f(x)=P(x)+\frac{C}{x-2}$, where $P(x)$ is a polynomial and $C$ a constant. The conclusion follows.
Another approach without mentioning poles(but uses the same idea): since the power series of $f(x)$ converges to $f(x)$ at least in $(0,2)$, it tends to infinity as $x\to 2$. Hence it cannot converge in an interval containing $2$. This shows its radius of convergence has to be $2$.