real analysis - Prove $lim_{nrightarrowinfty}x^{n}=0$

I would like to know if my proof is valid, because I did it different from the solution in my textbook (which uses Bernoulli's inequality).

If $|x|<1$, then $\lim_{n\rightarrow\infty}x^{n}=0$.

Proof

For $x=0$ it is trivial, so we suppose that $0<|x|<1$. Let $N>\dfrac{\log(x\varepsilon)}{\log(x)}$, then
\begin{align*}
|x|^{n}\end{align*}
implies that $\lim_{n\rightarrow\infty}x^{n}=0$ when $|x|<1$.

Answer

Suppose $\;0

$$x=\frac1r\;,\;\;1

Hint for the last part: assume it is false and use the archimedean property of the reals...