Suppose x lies in the interval (1, 3) with x≠2. Find the smallest positive value of δ such that the inequality 0<|x−2|<δ is true.

Verified step by step guidance

1. Understand the problem: We need to find the smallest positive value of δ such that the inequality 0 < |x−2| < δ holds true for x in the interval (1, 3) with x ≠ 2.
2. Consider the interval (1, 3) and identify the points where x is closest to 2. Since x ≠ 2, we need to consider the values of x approaching 2 from both sides within the interval.
3. Calculate the distance from 2 to the endpoints of the interval. The distance from 2 to 1 is |2 - 1| = 1, and the distance from 2 to 3 is |3 - 2| = 1.
4. Since x cannot be equal to 2, the smallest positive value of δ must be less than the smallest distance from 2 to the endpoints of the interval, which is 1.
5. Conclude that the smallest positive value of δ is slightly less than 1, ensuring that 0 < |x−2| < δ holds true for all x in the interval (1, 3) with x ≠ 2.

Key Concepts

Absolute Value Inequality

An absolute value inequality of the form |x - a| < δ describes the distance between x and a being less than δ. In this context, it means that x is within δ units of the point a (which is 2 here). Understanding this concept is crucial for determining the range of x values that satisfy the inequality.

Intervals

An interval is a set of real numbers that lie between two endpoints. The interval (1, 3) indicates that x can take any value greater than 1 and less than 3, excluding the endpoints. This concept is important for identifying the valid values of x that can be used in the inequality.

Limit and Continuity

In calculus, limits describe the behavior of a function as it approaches a certain point. The condition x ≠ 2 implies that we are examining the behavior of x as it approaches 2 from either side, which is essential for understanding how δ can be chosen to maintain the inequality while respecting the constraints of the interval.