Finding values of $x$ in the inequality

Question

We have this inequality for all $n \in \mathbb{N}$ : $\frac{3n-1}{n+1} \le |x-1| \le 3 + \frac{n}{n^2 + 1}$ . Find all of the possible values for $x$ .

My try : I tried to apply this lemma, but I didn't get any result : "If for all $h \gt 0 $ we have $0\le a \lt h$ then $a = 0$ ."

Answer 1

We need $$\sup_{n\in\mathbb N}\frac{3n-1}{n+1}\leq|x-1|\leq3+\inf_{n\in\mathbb N}\frac{n}{n^2+1}$$ or $$3\leq|x-1|\leq3,$$ which gives $x=4$ or $x=-2$.

Because if $|x-1|>3$ then there is $n\in\mathbb N$, for which $3+\frac{n}{n^2+1}<|x-3|$, which is contradiction with the given.

If $|x-1|<3$ then there is $n\in\mathbb N$, for which $\frac{3n-1}{n+1}>|x-1|,$ which is contradiction again.

Done!

Answer 2

doing case work we obtain: $$x\geq1 $$ and we have $$\frac{3n-1}{n+1}\le x-1\le 3+\frac{n}{n^2+1}$$ adding $1$ we have $$\frac{4n}{n+1}\le x\le 4+\frac{n}{n^2+1}$$ can you finish? for $n$ tend to infinity we get $$x=4$$ in the other case we get $$\frac{2(1-n)}{n+1}\geq x\geq \frac{-2n^2-n-2}{n^2+1}$$ therefore $$x=-2$$

Answer 3

Let $x$ such that

$\frac{3n-1}{n+1} \le |x-1| \le 3 + \frac{n}{n^2 + 1}$ for all $n$.

With $n \to \infty$ we get: $|x-1|=3$, hence $x=4$ or $x=-2$.