Question

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

Short answer

Answer: The smallest positive value of 𝛿 is 1.

Step-by-step solution

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 $x\ne 2$.

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 $x=1+\epsilon$, where $\epsilon$ represents a small positive value. 2. When $x$ is slightly less than 3, let's say $x=3-\epsilon$, where $\epsilon$ represents a small positive value.

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 $x=1+\epsilon$, we have $|x-2|=|(1+\epsilon -2)|=|(1-2+\epsilon )|=|(1-\epsilon )|.$ 2. For $x=3-\epsilon$, we have $|x-2|=|(3-\epsilon -2)|=|(3-2-\epsilon )|=|(1-\epsilon )|.$ Clearly, in both cases, the absolute value of the difference between $x$ and 2 is the same i.e., $|(1-\epsilon )|$.

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 $\epsilon$ is an infinitesimally small positive value. Hence, we can choose $\delta$ to be just larger than $(1-\epsilon ),$ which is also equal to $1$. So the smallest positive value of $\delta$ is $\boxed{{\displaystyle 1}}$.