Question

Using the \(\varepsilon-\delta\) Definition of Limit In Exercises \(37-48\) , find the limit \(L\) . Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L .\) $$ \lim _{x \rightarrow 3}|x-3| $$

Short answer

The limit of the function \(|x-3|\) as \(x\) approaches 3 is 0, as \(x\) approaches 3, the function \(|x-3|\) approaches 0. Moreover, we successfully proved this limit using the \(\varepsilon-\delta\) definition of limit.

Step-by-step solution

Determining the Limit

Using the definition of limit, we can substitute values of \(x\) closer to 3 in the function \(|x-3|\). As \(x\) approaches 3, \(|x-3|\) approaches 0. Therefore, \(L=0\).

Proving the Limit Using \(\varepsilon-\delta\)

We need to show for every \(\varepsilon>0\), there exists a \(\delta>0\) such that if \(0<|x-3|<\delta\), then \(|x-L|<\varepsilon\). Here, \(L=0\). Given \(\varepsilon>0\), let \(\delta=\varepsilon\). So, if \(0<|x-3|<\delta\), then we have \(0<|x-3|<\varepsilon\). Which means \(|x-L|<\varepsilon\). Therefore, the proof is complete.

Key concepts

Limit of Absolute Value Function

The concept of limits in calculus is crucial for understanding how functions behave as they approach different points. When dealing with the absolute value function, \(|x-c|\), it's essential to figure out what happens as \(x\) approaches a specific value, say 3 in this case. For the function \(|x-3|\), as \(x\) gets closer to 3, the expression inside the absolute value becomes smaller and smaller. Thus, \(|x-3|\) approaches 0. This is why the limit \(\lim\_{x \rightarrow 3}|x-3|\) is 0.

• The absolute value function measures the distance from a point on the real number line.
• When this distance approaches zero, the function's limit approaches zero too.

Recognizing this helps in understanding the behavior of not just absolute value functions but many forms of functions in calculus.

Proof Techniques in Calculus

The epsilon-delta (\(\varepsilon-\delta\)) definition is a formal way to prove limits. It can seem a bit abstract at first but is a powerful tool. Using this method, we show that we can make the function value as close as we like to a limit by controlling how close \(x\) gets to a specific value.

• \(\varepsilon\): Represents how close the function value is to the limit.
• \(\delta\): Represents how close \(x\) is to the point we're investigating.

For our example, we want to show that for every \(\varepsilon > 0\), we can find a corresponding \(\delta > 0\) such that if \(0 < |x - 3| < \delta\), then the inequality \(|x-3| < \varepsilon\) holds true. By setting \(\delta = \varepsilon\), we neatly show that the distance from the limit (which is 0) can be made as small as desired, proving the limit rigorously.

Calculus Problem Solving

Tackling calculus problems like proving limits using the epsilon-delta definition requires a strategic approach. Here's how you can systematically solve such problems:1. **Identify the Limit:** Start by determining what the limit is as \(x\) approaches the specified value. This often involves substituting values to observe the pattern.2. **Setup the Proof:** With \(\varepsilon-\delta\), create a logical framework that connects how close \(x\) is to the point to how close the function is to the limit.3. **Choose \(\delta\):** Pick a \(\delta\) value (often \(\varepsilon\)) that fulfills the condition \(0 < |x-c| < \delta\) leading to \(|f(x) - L| < \varepsilon\).4. **Verify and Conclude:** Make sure every step satisfies the definitions and concludes with the desired proof.By understanding these steps, students can gain confidence in problem-solving in calculus, reinforcing not just the understanding of limits but the logic and rigor necessary in more advanced topics.