"Let $x \in (0,1)$. Compute with careful proof the lowest upper bound and greatest lower bound of $x^n$, where $n \in\mathbb{N} $."
From my understanding, I would say the lowest upper bound is $1$. Clearly, $1$ is an upper bound of $x^n$ since numbers that are less than $1$ raised to a power greater than $1$ decrease in size. However, how would go about proving that it is indeed $1$?
As for the greatest lower bound, I would say it would be $0$. Clearly it is a lower bound since $0$ to a power is always $0$, which is less than the possible outcomes for $x$. However, I also remember that as n approaches infinity, the limit of $0^0$ approaches $1$, so that has me a bit confused on whether $0$ is the greatest lower bound I'm looking for.
Best Answer
It seems clear that $x$ is intended to be a fixed number in the interval $(0,1)$.
The answer to the least upper bound question depends on how one defines $\mathbb{N}$. If $0$ is included in $\mathbb{N}$, then the least upper bound is $1$. If $0$ is not included, then the least upper bound is $x$. That will be easy to prove, since $x^n\lt x^1$ for every $n\ge 2$. So the least upper bound is in fact a maximum.
The greatest lower bound is $0$, but not for the reason you described. Certainly $0\lt x^n$ for every $n\in \mathbb{N}$, which makes $0$ a lower bound. We want to show that there is no lower bound greater than $0$. So you need to show that for any $b\gt 0$, there is an $n$ such that $x^n \lt b$.
To do this, I suggest rewriting $x$ as $\frac{1}{1+y}$, where $y$ is positive. We want to show that for some $n$, we have $\frac{1}{(1+y)^n}\lt b$, or equivalently that there is an $n$ such that $(1+y)^n \gt 1/b$.
You can prove that result by observing that $(1 + y)^n \ge 1 + ny$ for $y \ge 0$ (It is true for $y \gt -1$.) This is the Bernoulli Inequality, which you may already have met. The Inequality can be proved by induction. Or else you can show using the Binomial Theorem that $(1 + y)^n \ge 1 + ny$.