Problem 1 - Calculus Early Transcendentals

Question

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.

Step-by-Step Solution

Answer

Answer: The smallest positive value of 𝛿 is 1.

Step 1: 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 \(x \neq 2\).

Step 2: 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.

Step 3: 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., \(|(1-\varepsilon)|\).

Step 4: 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 \(\boxed{1}\).