Answer
When $n \to \infty$ , then we have $|R_n(x)| \leq \dfrac{1}{(n+1)!} \to 0$.
Work Step by Step
The remainder in the Taylor series expansion of the function $f(x)$ centred at $a$ can be written as:
$|R_n(x)|=\dfrac{f^{n+1}(c)(x-a)^{n+1}}{(n+1)!}$
Now, we have the Taylor series expansion of the function $f(x)$ centred at $a=0$ as:
$f^{(n+1)}(c)=|\dfrac{(-1)^n}{(1+c)^{n+1}}$
and $|R_n(x)| \leq \dfrac{|x|^{n+1}}{(n+1)!}$ with $|f^{(n+1)}(c)|\leq 2^{n+1}|$ for $c \in (0,x)$ and $|x|^{n+1} \leq \dfrac{1}{2^{n+1}}$
Therefore, when $n \to \infty$ , then we have $|R_n(x)| \leq \dfrac{1}{(n+1)!} \to 0$.
