Using $\varepsilon$-analysis to prove the convergence of the sequence $(x^{n})$ for $|x| < 1$

Question

It is apparent that if $|x| < 1$ then $(x^{n})$ converges to $0$. Using $\varepsilon$-analysis to justify this matter is not that apparent to me, however. The problem I came across is to choose a suitable $N$.

Fix such an $x$ and let $\varepsilon > 0.$ For $n \geq N$ we have $|x^{n}| = |x|^{n} \leq |x|^{N},$ so to make $|x^{n}| < \varepsilon$, making $|x|^{N} \leq \varepsilon$ suffices. To proceed to choose such an $N$, I need an approach not involving logarithm.

Answer 1

For any fixed $\varepsilon >0$, you can choose $N$ to be the smallest positive integer bigger than or equal to $\frac{\log \varepsilon}{\log |x|}$, then for any $n>N \geq \frac{\log \varepsilon}{\log |x|}$, the inequality $|x|^n < \varepsilon$ should hold.

However this does not take into account $x=0$, you can consider it as a seperate case.

Essentially, we can work backwards from $|x|^n < \varepsilon$ to get the $N$.