In my previous question @SimpleArt came up with the wanted solution and I know that it always equals $\text{x}_\text{k}$ for $\text{m}\in\mathbb{N}^+$:
$$\text{x}_\text{k}=\frac{\text{k}!(-1)^\text{m}}{(\text{k}-\text{m})!}\cdot\text{x}_{\text{k}-\text{m}}+e\sum_{\text{p}=0}^{\text{m}-1}\frac{\text{k}!(-1)^\text{p}}{(\text{k}-\text{p})!}$$
Now, I want to find the limit, when $\text{m}\to\infty$ how do I do that:
$$\lim_{\text{m}\to\infty}\text{x}_\text{k}=\lim_{\text{m}\to\infty}\left\{\frac{\text{k}!(-1)^\text{m}}{(\text{k}-\text{m})!}\cdot\text{x}_{\text{k}-\text{m}}+e\sum_{\text{p}=0}^{\text{m}-1}\frac{\text{k}!(-1)^\text{p}}{(\text{k}-\text{p})!}\right\}$$
Since we have
$$x_k=\frac{k!(-1)^m}{(k-m)!}x_{k-m}+e\sum_{p=0}^{m-1}\frac{k!(-1)^p}{(k-p)!}$$
for all $m$, where $m$ and $k$ are independent variables, we have
$$\lim_{m\to\infty}x_k=x_k$$
and
$$\lim_{m\to\infty}\frac{k!(-1)^m}{(k-m)!}x_{k-m}+e\sum_{p=0}^{m-1}\frac{k!(-1)^p}{(k-p)!}=x_k$$