Answer
$\lim\limits_{n \to \infty} R_n(x)=0$
Work Step by Step
The Taylor series associated to $f$ is:
$f(x)\approx \sum_{k=0}^{\infty} \dfrac{f^{(k)}(x)}{k!}(x-a)^k$
Rewrite the series:
$f(x)\approx S_n(x)+R_n(x)$,
where $S_n(x)=\sum_{k=0}^{n} \dfrac{f^{(k)}(x)}{k!}(x-a)^k$
$R_n(x)=\sum_{k=n+1}^{\infty} \dfrac{f^{(k)}(x)}{k!}(x-a)^k$
As we are given that Taylor series converges to $f$, it means $\lim\limits_{n \to \infty} S_n(x)=f(x)$. We have:
$\lim\limits_{n \to \infty} f(x)=\lim\limits_{n \to \infty} S_n(x)+\lim\limits_{n \to \infty} R_n(x)$
$f(x)=f(x)+\lim\limits_{n \to \infty} R_n(x)$
$\lim\limits_{n \to \infty} R_n(x)=0$
