Below is the graph of the function f(x) = 1/x. find a positive number delta such that for all x satisfying 0 < | x - 1 | < delta, it holds that | f(x) - 1 | < 1/2. (if you prefer the string lingo, this question is essentially asking to find the width of the symmetric vertical strip around x = 1 such that for all x values from the vertical strip, the corresponding f(x) values lie in the horizontal strip shown in the graph. The width of this vertical strip is 2 times delta for some positive number delta, and you need to find delta.)

We are asked to find a positive number \( \delta \) such that for all \( x \) satisfying \( 0 < |x - 1| < \delta \), it holds that \( |f(x) - 1| < \frac{1}{2} \), where the function is \( f(x) = \frac{1}{x} \).

### Step-by-Step Solution:

We need to solve the inequality \( |f(x) - 1| < \frac{1}{2} \), or equivalently, \( \left| \frac{1}{x} - 1 \right| < \frac{1}{2} \).

1. **Rewrite the absolute value inequality**:
   \[
   \left| \frac{1}{x} - 1 \right| = \left| \frac{1 - x}{x} \right| < \frac{1}{2}
   \]
   This inequality simplifies to:
   \[
   \frac{|1 - x|}{|x|} < \frac{1}{2}
   \]
   
2. **Isolate \( |1 - x| \)**:
   Multiply both sides of the inequality by \( |x| \) (which is positive for \( x > 0 \)):
   \[
   |1 - x| < \frac{|x|}{2}
   \]
   
3. **Consider the neighborhood around \( x = 1 \)**:
   Since we are working around \( x = 1 \), we expect \( x \) to be close to 1. Let’s assume \( x \) is close enough to 1 so that \( \frac{1}{2} \leq x \leq \frac{3}{2} \). This is a reasonable assumption because for values of \( x \) far from 1, the inequality won't hold. So within this range:
   \[
   |x| \approx 1 \quad \text{(since \( x \) is close to 1)}
   \]
   
   Using this approximation, the inequality simplifies to:
   \[
   |1 - x| < \frac{1}{2}
   \]
   
4. **Find \( \delta \)**:
   The condition \( |1 - x| < \frac{1}{2} \) corresponds directly to \( \delta = \frac{1}{2} \), since \( |x - 1| < \delta \) describes the width of the vertical strip around \( x = 1 \).

Thus, the value of \( \delta \) is \( \frac{1}{2} \).

### Conclusion:

A positive number \( \delta \) that satisfies the condition \( | f(x) - 1 | < \frac{1}{2} \) for all \( x \) such that \( 0 < |x - 1| < \delta \) is \( \delta = \frac{1}{2} \).

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Do you have any questions or need further details?

Here are 5 related questions to consider:
1. How would the value of \( \delta \) change if we required \( |f(x) - 1| < \frac{1}{4} \)?
2. What does the function \( f(x) = \frac{1}{x} \) represent geometrically on a graph?
3. Can we find the \( \delta \) for other values of \( f(x) \), such as \( f(x) = \frac{1}{x^2} \)?
4. What would the behavior of \( f(x) \) be for large \( x \), and how does it affect \( \delta \)?
5. How does the choice of \( \delta \) reflect the continuity of \( f(x) \) near \( x = 1 \)?

**Tip**: Whenever dealing with inequalities involving functions, rewriting the inequality in simpler terms can greatly help to visualize and solve the problem.

Below is the graph of the function f(x) = 1/x. Find a positive number delta such that for all x satisfying 0 < |x - 1| < delta, it holds that |f(x) - 1| < 1/2. (This question is essentially asking to find the width of the symmetric vertical strip around x = 1 such that for all x values from the vertical strip, the corresponding f(x) values lie in the horizontal strip shown in the graph. The width of this vertical strip is 2 times delta for some positive number delta, and you need to find delta.)

Discover how to solve the limits problem for f(x) = 1/x by finding a positive delta such that |f(x) - 1| < 1/2. This detailed solution involves concepts from inequalities, limits, and the epsilon-delta definition of limits, suitable for high school students in Grades 11-12.

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