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.
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}
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...
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