Analytic Function on Neighborhood [closed]

We can write $f(x)= \sum\limits_{k=1}^{\infty} a_n(x-x_0)^{n}$ in some disk $D(x_0,R)$ containing $x_0$. Let $g(x)=\sum\limits_{k=1}^{\infty} a_n(x-x_0)^{n-1}$. Then $g$ is analytic on $V$ and $f(x)=(x-x_0)g(x)$ for all $x \in V$. If $U$ is the given neighborhood of $x_0$ Let $h(x)=g(x)$ for $x \in D(x_0,R/2)$ and $\frac {f(x)} {x-x_0}$ for $x \in U\setminus D(x_0,r/2)$. I will let you check that $h$ is analytic in $U$ and that $h(x)=\frac {f(x)} {x-x_0}$ for all $x \in U$.

Alternatively, you check that $h(x)=\frac {f(x)} {x-x_0}$ for $x \neq x_0$ and $f'(x_0)$ for $x=x_0$ defines a differentiable, hence analytic, function on $U$.