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Chapter 4: Problem 2

Determine a condition on \(|x-4|\) that will assure that: (a) \(|\sqrt{x}-2|<\frac{1}{2}\), (b) \(|\sqrt{x}-2|<10^{-2}\).

Short Answer

Expert verified
The condition on \(|x-4|\) that would be able to assure (a) \(|\sqrt{x}-2|<\frac{1}{2}\) is \(x < \frac{5}{4}\) and \(x > \frac{15}{4}\), and (b) \(|\sqrt{x}-2|<10^{-2}\) is \(x < 4 + 10^{-4}\) and \(x > 4 - 10^{-4}\). The common value for both conditions is \(x = 4\).

Step by step solution

01

Analyze the Absolute Value Inequalities - Part (a)

We start by analyzing the inequality \(|\sqrt{x}-2| < \frac{1}{2}\). First, solve the inequality by finding two possible expressions for \(|\sqrt{x}-2|\): either \(|\sqrt{x}-2| = \sqrt{x}-2\) or \(|\sqrt{x}-2| = 2-\sqrt{x}\), representing the cases where \(x\) is larger or smaller than \(2^2 = 4\).
02

Solve both inequalities - Part (a)

Now, to derive \(x\) from these, we square both sides of both equations to get \(x-4 < \frac{1}{4}\) and \(4 - x < \frac{1}{4}\). This gives us the solutions \(x < \frac{5}{4}\) and \(x > \frac{15}{4}\).
03

Discover the common value - Part (a)

We can see that the common value is \(x = 4\) for both conditions.
04

Analyze the Absolute Value Inequalities - Part (b)

Next, we analyze the inequality \( |\sqrt{x}-2| < 10^{-2}\). Following a pattern similar to that of Part (a), we find two possible expressions for \(|\sqrt{x}-2|\), either \(|\sqrt{x}-2| = \sqrt{x}-2\) or \(|\sqrt{x}-2| = 2-\sqrt{x}\), representing the cases where \(x\) is larger or smaller than \(2^2 = 4\).
05

Solve both Inequalities - Part (b)

Again, to derive \(x\) from these, we square both sides of both equations to get \(x-4 < 10^{-4}\) and \(4 - x < 10^{-4}\). This gives us the solutions \(x < 4 + 10^{-4}\) and \(x > 4 - 10^{-4}\).
06

Discover the common value - Part (b)

We can see that the common value is \(x = 4\) for both conditions.

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Most popular questions from this chapter

Give examples of functions \(f\) and \(g\) such that \(f\) and \(g\) do not have limits at a point \(c\), but such that both \(f+g\) and \(f g\) have limits at \(c\).

Let \(f, g\) be defined on \(A\) to \(\mathbb{R}\) and let \(c\) be a cluster point of \(A\). (a) Show that if both \(\lim _{x \rightarrow c} f\) and \(\lim _{x \rightarrow c}(f+g)\) cxist, then \(\lim _{x \rightarrow c} g\) exists. (b) If \(\lim _{x \rightarrow c} f\) and \(\lim _{x \rightarrow c} f g\) exist, does it follow that \(\lim _{x \rightarrow c} g\) exists?

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Suppose that \(\lim _{x \rightarrow c} f(x)=L\) where \(L>0\), and that \(\lim _{x \rightarrow c} g(x)=\infty\). Show that \(\lim _{x \rightarrow c} f(x) g(x)=\) \(\infty\). If \(L=0\), show by example that this conclusion may fail.

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