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The problem is asking whether the function \( f(x) = \frac{1}{x} \), defined on the interval \( (a, \infty) \) with \( a > 0 \), is uniformly continuous. Let’s examine this step by step.

### Definition of Uniform Continuity:
A function \( f \) is uniformly continuous on a domain if, for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x_1, x_2 \in (a, \infty) \), if \( |x_1 - x_2| < \delta \), then \( |f(x_1) - f(x_2)| < \epsilon \).

### Steps:
1. The function given is \( f(x) = \frac{1}{x} \), which is continuous on \( (a, \infty) \).
2. For uniform continuity, we need to check whether the function satisfies the definition above for all points in the interval \( (a, \infty) \).

### Argument for Uniform Continuity:
We calculate the difference:
\[
|f(x_1) - f(x_2)| = \left| \frac{1}{x_1} - \frac{1}{x_2} \right| = \left| \frac{x_2 - x_1}{x_1 x_2} \right|
\]
Now, since \( x_1, x_2 > a \) (where \( a > 0 \)), both \( x_1 \) and \( x_2 \) are bounded below by \( a \), meaning \( x_1 x_2 \geq a^2 \).

Thus,
\[
|f(x_1) - f(x_2)| \leq \frac{|x_1 - x_2|}{a^2}
\]
This shows that if \( |x_1 - x_2| < \delta \), we can make \( |f(x_1) - f(x_2)| \) smaller than any \( \epsilon \) by choosing \( \delta = a^2 \epsilon \).

### Conclusion:
Yes, the function \( f(x) = \frac{1}{x} \) is uniformly continuous on \( (a, \infty) \) for any \( a > 0 \). This is because the difference \( |f(x_1) - f(x_2)| \) is proportional to \( |x_1 - x_2| \), and we can always find a \( \delta \) that satisfies the condition for uniform continuity.

Would you like more details or have any questions?

### Relative Questions:
1. What is the difference between uniform continuity and pointwise continuity?
2. How does the function behave as \( x \to \infty \)?
3. What happens if the interval is extended to include 0?
4. Can you provide an example of a function that is continuous but not uniformly continuous?
5. How would you prove uniform continuity using the definition for other types of functions?

### Tip:
When checking for uniform continuity, always ensure that the behavior of the function is controlled over the entire domain, not just locally.

For a > 0 real number, let f: (a, ∞) → R be a function defined as f(x) = 1/x. Can we say that f is a uniformly continuous function? Justify your answer.

Explore the proof of uniform continuity for the function f(x) = 1/x, defined on the interval (a, ∞) where a > 0. This advanced calculus problem delves into the uniform continuity definition, leveraging the difference quotient to show that f(x) is uniformly continuous.

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