Understanding a proof that if $|x|<1$ then $x^n \to 0$

Question

The question:

The answer:

I understand the answer of $(d)$ but I don't know how to put "$\alpha= 0$" in order to get the solution of $(e)$.

Help?

Answer 1

If $x=0$, it is trivial. Otherwise, asserting that $\lim_{n\to\infty}x^n$ is equivalent to asserting that $\lim_{n\to\infty}|x|^n=0$. But$$|x|^n=\frac1{(1/|x|)^n}.$$So, take $p=\frac1{|x|}-1$.

Answer 2

Put $\alpha=0$ so that part (d) tells us that $$\lim_{n\to\infty}\frac{n^0}{(1+p)^n}=\lim_{n\to\infty}\frac{1}{(1+p)^n}=0$$ then since $p>0$ we have $0<\frac{1}{1+p}<1$

Answer 3

Note that for $\alpha=0$ the last inequality becomes

$$0<\frac{1}{(1+p)^n}<\frac{2^kk!}{p^kn^k}$$

Answer 4

The hint is trying to say:

Set $\alpha=0$ and $x=\frac{1}{p+1}$ in (d). Note that $x= \frac{1}{p+1} \Leftrightarrow p = \frac{1}{x}-1>0$.