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Chapter 5: Problem 1

Show that the function \(f(x):=1 / x\) is uniformly continuous on the set \(A:=[a, \infty)\), where \(a\) is a positive constant.

Short Answer

Expert verified
Yes, the function \(f(x) = 1/x\) is uniformly continuous on the set \(A=[a, \infty)\). The function meets the requirement for uniform continuity that for every \(\epsilon > 0 \), there exists a \(\delta = \epsilon a^2 > 0\) such that if \(|x-y| < \delta\) for any \(x, y\) in set \(A\), then \(|f(x) - f(y)| < \epsilon\).

Step by step solution

01

Recall the Definition of Uniform Continuity

Uniform continuity is defined as: if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x-y| < \delta\) for any \(x, y\) in a set, then \(|f(x) - f(y)| < \epsilon\). Our task is to find such a \(\delta > 0\) for our function \(f(x) = 1/x\) and the set \(A=[a, \infty)\).
02

Determine the Difference \(|f(x) - f(y)|\)

For the function \(f(x) = 1/x\), the difference \(|f(x) - f(y)| = |1/x - 1/y|\). This can be simplified by combining the fractions to \(|(y-x)/xy|\). Given \(x, y \in A\), where \(A=[a, \infty)\), we know that \(x, y \geq a\). Therefore, \(xy \geq a^2\). Then, \(|(y-x)/xy| \leq |(y-x)/a^2|\).
03

Proof that a Suitable \(\delta\) Exists

We want \(|(y-x)/a^2| < \epsilon\), or equivalently, \(|y-x| < \epsilon a^2\). We can make this the definition of our \(\delta\) (i.e. \(\delta = \epsilon a^2\)). Hence, for any \(x, y \in A\) where \(|x-y| < \delta\), it follows that \(|f(x) - f(y)| < \epsilon\). This confirms that the function \(f(x) = 1/x\) is indeed uniformly continuous on the set \(A\).

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