Q. 6 - Calculus

Question

Why do we have $0<|x-c|<\delta$ instead of just $|x-c|<\delta$ in Definition 1.10? 

Step-by-Step Solution

Answer

As the absolute value of a number is always positive or zero, it is always equivalent to saying that $|x-c|>0$

Step 1. Given Information

The given statement is we have  $0<|x-c|<\delta$ instead of just $|x-c|<\delta$ in Definition 1.10

Step 2. Explanation

Definition 1.10 states that- The limit expression $\underset{x\to c}{lim}f\left(x\right)=L$ means that for all epsilon positive there exists delta positive such that if $0<|x-c|<\delta \mathrm{then} \left|f\right(x)-L|<\epsilon$

In this, the punctured interval is given by $x\in (c-\delta ,c)\cup (c,c+\delta )$ which means that $c-\delta <x<c+\delta$

Now subtract c from each part,

$$\begin{gathered} c-\delta -c<x-c<c+\delta -c \\ -\delta <x-c<\delta \end{gathered}$$

This is the solution set for the inequality $|x-c|<\delta$

The fact that $x\ne c$means that $x-c\ne 0$ which is equivalent to saying that $\left|x-c\right|\ne 0$ and since the absolute value of a number is always positive or zero, it is always equivalent to saying that $|x-c|>0$

Thus, we must write $0<|x-c|<\delta$ but not only $|x-c|<\delta$