Question on the convergence of $|x|^n\rightarrow 0$ when $|x|<1$.

Question

My question is about calculating the $N$ for a convergent sequence.

I want to prove that $|x|^n \rightarrow 0$ when $|x|<1$ using the $\epsilon, N$ definition.

So I know that $|x^n-0| \leq |x|^n$ and I want $|x|^n<\epsilon$.

Such an $N$ was already provided $N = \max(1,\log_{|x|}(\frac{\epsilon}{2})).$

I understand exactly why $N = \log_{|x|}(\frac{\epsilon}{2})$ works.

My questions are :

1) Why choose from the $\max$ of these two? In particular why 1? This is the main this I want to understand.

2) How to calculate such an $N.$ I performed:

$$ n\log(|x|)\leq\log(\frac{\epsilon}{2}) \implies n\geq\frac{\log(\frac{\epsilon}{2})}{\log(|x|)} = \log_{|x|}(\frac{\epsilon}{2})?$$ Is this correct? Any help and comments would be appreciated! Thank you.

Answer 1

You don't have to do anything from there. Given an $x$ and $\epsilon$, this gives an $n$ for which $|x|^n < \epsilon$.

Your only error is forgetting that $\log(|x|)$ and $\log(\epsilon)$ are both negative, so that the inequality should be $n\log(|x|)<\log(\epsilon) \implies n>\frac{\log(\epsilon)}{\log(|x|)} $.