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Chapter 2: Problem 1

Suppose \(x\) lies in the interval (1,3) with \(x \neq 2 .\) Find the smallest positive value of \(\delta\) such that the inequality \(0<|x-2|<\delta\) is true.

Short Answer

Expert verified
Answer: The smallest positive value of 饾浛 is 1.

Step by step solution

01

Understand the inequality and given condition

The inequality given to us is \(0<|x-2|<\delta\). This means the absolute value of \((x-2)\) should be positive and less than \(\delta\). We also know that \(x\) lies in the open interval (1,3) which implies that \(1 < x < 3\). Finally, we should note that \(x \neq 2\).
02

Determine the possible values of \(x\) within the interval

Since \(x\) cannot be equal to 2, we need to find the largest absolute difference between 2 and the remaining possible values of \(x\) in the interval. We have two possible cases to consider. 1. When \(x\) is slightly greater than 1, let's say \(x=1+\varepsilon\), where \(\varepsilon\) represents a small positive value. 2. When \(x\) is slightly less than 3, let's say \(x=3-\varepsilon\), where \(\varepsilon\) represents a small positive value.
03

Calculate the absolute value of the difference between \(x\) and 2

Now, we need to calculate the absolute value of the difference between \(x\) and 2 for both cases. 1. For \(x=1+\varepsilon\), we have \(|x-2|=|(1+\varepsilon-2)|=|(1-2+\varepsilon)|=|(1-\varepsilon)|.\) 2. For \(x=3-\varepsilon\), we have \(|x-2|=|(3-\varepsilon-2)|=|(3-2-\varepsilon)|=|(1-\varepsilon)|.\) Clearly, in both cases, the absolute value of the difference between \(x\) and 2 is the same i.e., \(|(1-\varepsilon)|\).
04

Find the smallest value of \(\delta\)

Since we want the smallest positive value of \(\delta\) such that the inequality \(0<|x-2|<\delta\) is true, we should choose the \(\delta\) to be just greater than the smallest possible absolute value of the difference between \(x\) and 2 in both cases. In our case, the smallest possible value would be when \(\varepsilon\) is an infinitesimally small positive value. Hence, we can choose \(\delta\) to be just larger than \((1-\varepsilon),\) which is also equal to \(1\). So the smallest positive value of \(\delta\) is \(\boxed{1}\).

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