How to prove a limit? I saw at the solution of an exercise that $$\frac{|x|^{n+1}}{(n+1)!} \to 0, \text{ when n  } \to +\infty$$
But,how can I show that it is actually like that?
 A: Take $x$ positive, that does no harm. Now consider $n \to \infty$ in:
$$
\frac{x^{n + 2} / (n + 2)!}{x^{n + 1} / (n + 1)!}
  = \frac{x}{n + 2} \to 0
$$
As the ratio tends to zero, so does each term.
A: Let $k\in \mathbb{N}$ such that $k + 1 > \vert x \vert$. For $n>k$, we can write the term as
$$\frac{\vert x \vert \cdot \, \dots \, \cdot \vert x \vert\cdot \vert x \vert}{1\cdot \dots \, \cdot n \cdot (n+1)}= \frac{\vert x \vert^k }{1\cdot \dots \, \cdot k}\cdot \frac{\vert x \vert^{n-k}}{(k+1)\cdot \dots \, \cdot (n+1)}$$
The first ratio in this product is some number which does not depend on $n$. The second ratio has $(n-k)$ factors in the numerator and the denominator, thus it can be written as a product of $(n-k)$ ratios:
$$\frac{\vert x \vert^{n-k}}{(k+1)\cdot \dots \, \cdot (n+1)}= \frac{\vert x \vert}{k+1}\cdot\frac{\vert x \vert}{k+2}\cdot \dots \, \cdot \frac{\vert x \vert}{n+1},$$
where each of the ratios is less than $1$ (since $k+1 > \vert x \vert$). As $n$ increases, these ratios tend to zero.
