# $1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. 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: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$

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What happens with your previous questions which received answers? Do you plan to accept some of these? – Did Aug 22 '12 at 10:48
What have you tried? Is this homework? – Jeel Shah Aug 22 '12 at 10:49
The question is unclear. Do you mean that $1-\frac1n<x\le 3+\frac1n$ for all $n\in\Bbb N$? For at least one $n\in\Bbb N$? – Brian M. Scott Aug 22 '12 at 10:50
For each $n$ the range will vary. Are you trying, instead, to find out what range of $x$ will satisfy that for every natural number? – Pedro Tamaroff Aug 23 '12 at 1:11

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 $n$.

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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 $\inf\{s_n\}\notin \{s_n\}$.

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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 $[1,3]$.

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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, $0 \lt x \le 4$

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Wrong. $4\nleq 3 + \frac{1}{2}$ – user5137 Aug 23 '12 at 21:08