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Chapter 1: Q. 6 (page 107)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given Information  

The given statement is we have 0<|x-c|<δinstead of just |x-c|<δin Definition 1.10

02

Step 2. Explanation

Definition 1.10 states that- The limit expression limx→cf(x)=Lmeans that for all epsilon positive there exists delta positive such that if 0<|x-c|<δthen|f(x)-L|<ε

In this, the punctured interval is given by x∈(c-δ,c)∪(c,c+δ)which means that c-δ<x<c+δ

Now subtract c from each part,

c-δ-c<x-c<c+δ-c-δ<x-c<δ

This is the solution set for the inequality |x-c|<δ

The fact thatx≠cmeans thatx-c≠0which is equivalent to saying thatx-c≠0and sincethe absolute value of a number is always positive or zero, it is always equivalent to saying that |x-c|>0

Thus, we must write0<|x-c|<δbut not only|x-c|<δ

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Most popular questions from this chapter

Write each of the inequalities in interval notation:

0<|x-2|<0.1

For each limit statement , use algebra to find δ > 0 in terms of ε> 0 so that if 0 < |x − c| < δ, then | f(x) − L| < ε.

limx→0(5x2-1)=-1

In Exercises 39–44, use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each discontinuity of f, describe the type of discontinuity and any one-sided discontinuity.

f(x)=-x,ifx<02,ifx=0x,ifx>0.

For each of the following sign charts, sketch the graph of a function f that has the indicated signs, zeros, and discontinuities:

In Exercises 39–44, use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each discontinuity of f, describe the type of discontinuity and any one-sided discontinuity.

f(x)=x2,ifx<24,ifx=22x+1,ifx>2.
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