Problem 30 - Single Variable Calculus

Question

Use the definition in Exercise 28 to prove that \(\lim _{x \rightarrow 2^{+}} \sqrt{x-2}=0\)

Step-by-Step Solution

Answer

Using the definition of a limit, we want to prove that \(\lim_{x\rightarrow 2^+} \sqrt{x-2} = 0\). We can rewrite the conditions as \(0 < x - 2 < \delta\) and \(\sqrt{x-2} < \varepsilon\). We can find a relationship between \(\delta\) and \(\varepsilon\) by squaring the second condition: \(x - 2 < \varepsilon^2\). Then, we choose \(\delta = \varepsilon^2\) and verify that it satisfies the conditions from the definition. Therefore, we can conclude that \(\lim_{x \rightarrow 2^+} \sqrt{x-2} = 0\).

Step 1: Relate the given function to the definition

First, let's rewrite the conditions from the definition of a limit using our function and limit: - \(0 < x - 2 < \delta\) - \(|\sqrt{x-2} - 0| < \varepsilon\) Since \(\sqrt{x-2} \ge 0\), we can simplify the second condition to: - \(\sqrt{x-2} < \varepsilon\) Now our goal is to find an appropriate \(\delta\) that satisfies these conditions given an arbitrary \(\varepsilon > 0\).

Step 2: Find a relationship between \(\delta\) and \(\varepsilon\)

We want to find a relationship between \(\delta\) and \(\varepsilon\) that will satisfy the conditions from the definition of a limit. - Given \(\varepsilon > 0\), we want to find \(\delta > 0\) - If \(0 < x - 2 < \delta\), then \(\sqrt{x-2} < \varepsilon\) Let's square both sides of the inequality \(\sqrt{x-2} < \varepsilon\): - \((\sqrt{x-2})^2 < \varepsilon^2\) - \(x-2 < \varepsilon^2\) Now we can choose \(\delta\) as \(\delta = \varepsilon^2\).

Step 3: Verify that the chosen \(\delta\) satisfies the definition

For every \(\varepsilon > 0\), we have \(\delta = \varepsilon^2\). Let's check the conditions from the definition of a limit: - \(0 < x - 2 < \delta\), which means \(0 < x - 2 < \varepsilon^2\) - From the last step, we derived that \(x - 2 < \varepsilon^2\), so squaring both sides and taking square roots gives \(\sqrt{x-2} < \varepsilon\), as required. Hence, we have shown that for every \(\varepsilon > 0\), there exists a \(\delta > 0\), specifically \(\delta = \varepsilon^2\), such that if \(0 < x - 2 < \delta\), then \(\sqrt{x-2} < \varepsilon\). Thus, we can conclude that: \(\lim_{x \rightarrow 2^+} \sqrt{x-2} = 0\)

Key Concepts

Epsilon-Delta Definition of a Limit

Understanding the concept of a limit is a foundational element in calculus, and the epsilon-delta definition is a formal method to define what exactly a limit is. This definition tells us that a function \(f(x)\) approaches a limit \(L\) as \(x\) approaches a point \(a\) (from either side) if for every positive number \(\varepsilon\) (however small), there exists some positive number \(\delta\) such that whenever \(0 < |x - a| < \delta\), the value of the function \(f(x)\) will be within an \(\varepsilon\)-radius of \(L\), or \( |f(x) - L| < \varepsilon\).

The epsilon-delta definition puts a rigorous groundwork for understanding limits, ensuring that the function's value gets arbitrarily close to the limit. By choosing \(\varepsilon\) to represent any measure of closeness we desire, we are forced to find a \(\delta\) that guarantees \(f(x)\) is within this range of our limit \(L\). This precision enables mathematicians to discuss and prove the behavior of functions near specific points with clear-cut criteria.

Proving Limits

When proving limits using the epsilon-delta definition, we demonstrate that no matter how small the distance \(\varepsilon\) is from the limit \(L\), we can always find a range \(\delta\) around the point \(a\) such that the function's value falls within that \(\varepsilon\)-range. It's like playing a game where \(\varepsilon\) is how close you need to get to the target, and \(\delta\) is how close you need to aim your arrow near the bullseye \(a\) to ensure it lands within the \(\varepsilon\)-circle around the target \(L\).

In the given exercise, the aim was to prove that the function \(\sqrt{x-2}\) approaches 0 as \(x\) approaches 2 from the right. By manipulating the conditions and finding a relationship between \(\delta\) and \(\varepsilon\), and subsequently verifying that this relationship holds, we accomplish the proof. Such proofs are fundamental in calculus because they validate the intuitions about a function's behavior with mathematical rigor.

Right-Hand Limit

A right-hand limit takes into consideration the behavior of a function \(f(x)\) as \(x\) approaches a particular value \(a\) from the right, meaning values greater than \(a\). Symbolically, this is notated as \(\lim_{x \rightarrow a^+} f(x)\). The '+' superscript indicates the direction from which \(x\) is approaching \(a\).

For instance, the example problem demonstrates a right-hand limit where \(x\) approaches the value of 2 from the right. This is crucial because the behavior of \(f(x)\) on the right side of \(a\) might differ from its behavior on the left, especially at points where the function is not continuous or has a discontinuity. Establishing the existence (or non-existence) of right-hand (and left-hand) limits can thus reveal much about the nature of functions at specific points. Considering only one side of the approach also simplifies the \(\delta-\varepsilon\) game, because we only need to worry about values of \(x\) in a certain direction relative to \(a\).