A real number x satisfying `1-(1)/(n) lt x le 3 + (1)/(n)`, for every positive integer n, is best described by

Options

A. `1 lt x lt 4`
B. `1 lt x le 3`
C. `0 lt x le 4`
D. `1 le x le 3`

Answer

To solve the inequality \( 1 - \frac{1}{n} < x \leq 3 + \frac{1}{n} \) for every positive integer \( n \), we will analyze the bounds as \( n \) varies.

### Step 1: Analyze the lower bound
The lower bound of the inequality is given by:
\[ 1 - \frac{1}{n} \]
As \( n \) increases (i.e., \( n = 1, 2, 3, \ldots \)), the term \( \frac{1}{n} \) decreases, which means \( 1 - \frac{1}{n} \) increases.

- For \( n = 1 \):
\[ 1 - \frac{1}{1} = 0 \]
- For \( n = 2 \):
\[ 1 - \frac{1}{2} = 0.5 \]
- For \( n = 3 \):
\[ 1 - \frac{1}{3} \approx 0.67 \]

As \( n \) approaches infinity, \( 1 - \frac{1}{n} \) approaches \( 1 \). Therefore, the lower bound approaches \( 1 \) but never reaches it.

### Step 2: Analyze the upper bound
The upper bound of the inequality is given by:
\[ 3 + \frac{1}{n} \]
As \( n \) increases, the term \( \frac{1}{n} \) also decreases, which means \( 3 + \frac{1}{n} \) decreases.

- For \( n = 1 \):
\[ 3 + \frac{1}{1} = 4 \]
- For \( n = 2 \):
\[ 3 + \frac{1}{2} = 3.5 \]
- For \( n = 3 \):
\[ 3 + \frac{1}{3} \approx 3.33 \]

As \( n \) approaches infinity, \( 3 + \frac{1}{n} \) approaches \( 3 \). Therefore, the upper bound approaches \( 3 \) but includes \( 3 \).

### Step 3: Combine the bounds
From the analysis, we have:
- The lower bound approaches \( 1 \) (but is always less than \( 1 \)).
- The upper bound approaches \( 3 \) (and is equal to \( 3 \)).

Thus, we can conclude that:
\[ 1 < x \leq 4 \]

### Conclusion
The real number \( x \) satisfying \( 1 - \frac{1}{n} < x \leq 3 + \frac{1}{n} \) for every positive integer \( n \) is best described by:
\[ 1 < x \leq 4 \]