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Chapter 2: Problem 4

Does the set \(\\{x: 0<|x-a|<\delta\\}\) include the point \(x=a ?\) Explain.

Short Answer

Expert verified
Provide an explanation. Answer: No, the point \(x=a\) is not included in the set defined by the inequality \(0<|x-a|<\delta\) because when \(x=a\), the absolute value expression evaluates to 0, but the inequality requires it to be strictly greater than 0.

Step by step solution

01

Understand the given inequality

The inequality we are dealing with is \(0<|x-a|<\delta\). Keep in mind that \(\delta\) is a positive number, which defines an open interval around the point \(a\).
02

Analyze the absolute value

Absolute values are always non-negative, i.e., \(|x-a| \geq 0\) for any values of \(x\) and \(a\). However, our inequality states that \(|x-a|\) must be strictly greater than 0.
03

Check if \(x=a\) satisfies the inequality

We need to verify if the inequality holds when \(x=a\). In that case, we'll have an absolute value expression of \(|a-a|\). Since \(a-a=0\), the expression turns into \(|0|=0\).
04

Compare the result with the inequality

From Step 3, we find that when \(x=a\), the value of the absolute value expression is exactly 0. However, according to the inequality, we need \(|x-a|\) to be strictly greater than 0.
05

Determine if the point \(x=a\) is included in the set

Since the inequality states that \(|x-a|\) must be strictly greater than 0, and we found that \(|x-a|=0\) when \(x=a\), the point \(x=a\) is not included in the set defined by the inequality \(0<|x-a|<\delta\).

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Most popular questions from this chapter

Prove the following statements to establish the fact that \(\lim _{x \rightarrow a} f(x)=L\) if and only if \(\lim _{x \rightarrow a^{-}} f(x)=L\) and \(\lim _{x \rightarrow a^{+}} f(x)=L\) a. If \(\lim _{x \rightarrow a^{-}} f(x)=L\) and \(\lim _{x \rightarrow a^{+}} f(x)=L,\) then \(\lim _{x \rightarrow a} f(x)=L\) b. If \(\lim _{x \rightarrow a} f(x)=L,\) then \(\lim _{x \rightarrow a^{-}} f(x)=L\) and \(\lim _{x \rightarrow a^{+}} f(x)=L\)

a. Sketch the graph of a function that is not continuous at \(1,\) but is defined at 1. b. Sketch the graph of a function that is not continuous at \(1,\) but has a limit at 1.

Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\). Then state the horizontal asymptote(s) of \(f\). Confirm your findings by plotting \(f\). $$f(x)=\frac{3 e^{x}+e^{-x}}{e^{x}+e^{-x}}$$

Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. $$h(x)=\frac{e^{x}}{(x+1)^{3}}$$

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