$1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$. $1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$.
NOTE: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$
 A: If you are looking for the set of all $x$ which satisfy your inequalities for all $n\in\mathbb{N}$, then the answer is $[1,3]$, as can easily be seen by considering any value not in the interval $[1,3]$, and checking that it fails one of your inequalities for some value of $n$.
A: If $t_n<x<s_n$ for all $n\in \Bbb N$, where $t_n$ and $s_n$ are terms of sequences $\{t_n\}$ and $\{s_n\}$ respectively, then range of $x$ is $[\sup\{t_n\},\inf\{s_n\}]$ if $\sup\{t_n\}\notin \{t_n\}$ and $\inf\{s_n\}\notin \{s_n\}$.
A: I would look at it by splitting it into two inequalities. So for all natural numbers $n$ you need
$$1 - \frac{1}{n} < x.
$$
The solution to this one inequality is clearly $1 \leq x$. The other part says that for all natural numbers $n$ you need:
$$ x \leq 3 + \frac{1}{n}.
$$
The solution to this one inequality if clearly that $x \leq 3$.
Combining the two, you finally get that the solution is all $x$ in the interval $1 \leq x \leq 3$ or you can write it as an interval $[1,3]$.
A: when $n=1$,
$1-\frac{1}{n}=0$, $3+\frac{1}{n}=4$
when $n=1$, we have, $0 \lt x_{1} \le 4$
when $n=2$, we have, $\frac{1}{2} \lt x_{2} \le 3+\frac{1}{2}$
when $n \to \infty, \frac{1}{n} \to 0$, and then we have, $1 \lt x_{\infty} \le 3$
we can now conclude that, $0 \lt x \le 4$
