Question

Explain why $\underset{n\to \infty }{\lim }\frac{{\left|x\right|}^n}{n!}=0$for every value of x.

Short answer

Using ratio test of convergence,

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{{\left|x\right|}^{n+1}}{n+1!}\times \frac{n!}{{\left|x\right|}^n}=\underset{n\to \infty }{\lim }\left|x\right|\frac{n}{n+1} \\ =0 \end{gathered}$$

Step-by-step solution

Step 1. Given information.

The limit is$\underset{n\to \infty }{\lim }\frac{{\left|x\right|}^n}{n!}=0$.

Step 2. Explanation.

Use the ratio test of convergence.

Let $b_n=\frac{{\left|x\right|}^n}{n!}$and $b_{k+1}=\frac{{\left|x\right|}^{n+1}}{n+1!}$

role="math" localid="1649610457627" $$\begin{gathered} \underset{n\to \infty }{\lim }\frac{{\left|x\right|}^{n+1}}{n+1!}\times \frac{n!}{{\left|x\right|}^n}=\underset{n\to \infty }{\lim }\left|x\right|\frac{n}{n+1} \\ =0 \end{gathered}$$