$1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$.
NOTE: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$
$1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$.
NOTE: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$
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 $n$.
If $t_n
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 $[1,3]$.
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, $0 \lt x \le 4$