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A real number $x$ satisfying $1- \frac{1}{n} < x \leq 3 + \frac{1}{n}$ for every positive integer $n,$ is best described by

1. $1 < x < 4$
2. $0 < x \leq 4$
3. $0 < x \geq 4$
4. $1 \leq x \leq 3$

If n = 1, then 0 < x $\leq$ 4

when the value of n increases from 1 to infinity, the range of x approaches (1 < x $\leq$ 3).

So, the lower limit of x increases from almost 0 (to almost 1), while the higher limit decreases from 4 (to 3).

Thus, the lowest possible limit of x will be > 0 and the highest possible limit will be &le; 4.

For all range of n, 0 < x $\leq$ 4

Hence,Option(B) 0 < x $\leq$ 4.
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