[proofplan] We prove continuity at each point $x_0 \in (a-R, a+R)$ by localising to a compact sub-interval. The [Radius of Convergence theorem](/theorems/202) provides absolute uniform convergence on every closed sub-interval $[a-r, a+r]$ with $r < R$. Each partial sum is a polynomial and hence continuous, so the [uniform limit theorem](/theorems/258) delivers continuity of the sum on the sub-interval, and in particular at $x_0$. [/proofplan] [step:Localise $x_0$ inside a compact sub-interval where the series converges uniformly] Fix $x_0 \in (a-R, a+R)$. Choose $r$ with $|x_0 - a| < r < R$. By the [Radius of Convergence theorem](/theorems/202) (part 3), the power series $\sum_{n=0}^\infty c_n(x-a)^n$ converges absolutely and uniformly on $[a-r, a+r]$. Since $|x_0 - a| < r$, the point $x_0$ lies in the interior of $[a-r, a+r]$. [/step] [step:Apply the uniform limit theorem to obtain continuity at $x_0$] On $[a-r, a+r]$, each partial sum $S_N(x) = \sum_{n=0}^N c_n(x-a)^n$ is a polynomial in $x$, hence continuous. The sequence $(S_N)$ converges uniformly to $f$ on $[a-r, a+r]$ by the previous step. By the [uniform limit theorem](/theorems/258), the uniform limit of continuous functions is continuous, so $f$ is continuous on $[a-r, a+r]$. In particular, $f$ is continuous at $x_0$. [guided] We have localised $x_0$ to the compact interval $[a-r, a+r]$ on which the power series converges uniformly. The remaining task is to invoke the uniform limit theorem, but we must carefully verify its two hypotheses. **Hypothesis 1 (continuous approximants):** Each partial sum $S_N(x) = \sum_{n=0}^N c_n(x-a)^n$ is a polynomial of degree $N$. Polynomials are continuous on all of $\mathbb{R}$, so in particular each $S_N$ is continuous on $[a-r, a+r]$. **Hypothesis 2 (uniform convergence):** The sequence $(S_N)$ converges uniformly to $f$ on $[a-r, a+r]$. This follows from the [Radius of Convergence theorem](/theorems/202) (part 3), which guarantees absolute and uniform convergence on compact sub-intervals $[a-r, a+r]$ with $r < R$. With both hypotheses verified, the [uniform limit theorem](/theorems/258) gives the conclusion: the uniform limit of continuous functions is continuous. Therefore $f$ is continuous on $[a-r, a+r]$, and in particular at $x_0$. Why did we need to localise? The power series need not converge uniformly on all of $(a-R, a+R)$ --- uniform convergence is only guaranteed on compact sub-intervals $[a-r, a+r]$ with $r < R$. But continuity is a local property: to prove $f$ is continuous at $x_0$, it suffices to prove $f$ is continuous on any neighbourhood of $x_0$, and the interval $[a-r, a+r]$ (with $|x_0 - a| < r < R$) provides such a neighbourhood. [/guided] [/step] [step:Conclude continuity on the full interval by the arbitrary choice of $x_0$] Since $x_0 \in (a-R, a+R)$ was arbitrary, $f$ is continuous at every point of $(a-R, a+R)$, and hence $f$ is continuous on $(a-R, a+R)$. [/step]