Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$.

NOTE: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$

share|cite|improve this question
What happens with your previous questions which received answers? Do you plan to accept some of these? – Did Aug 22 '12 at 10:48
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 $n$.

share|cite|improve this answer

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

share|cite|improve this answer

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 $[1,3]$.

share|cite|improve this answer

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, $0 \lt x \le 4$

share|cite|improve this answer
Wrong. $4\nleq 3 + \frac{1}{2}$ – user5137 Aug 23 '12 at 21:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.