Problem 37 - Calculus of a Single Variable

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 \(L .\) $$ \lim _{x \rightarrow 4}(x+2) $$

Step-by-Step Solution

Answer

The limit of \(f(x) = x+2\) as \(x\) approaches 4 is \(L = 6\), and it has been proved using the \(\varepsilon-\delta\) definition of limit.

Step 1: Finding the Limit

To find the limit of \(f(x) = x+2\) as \(x\) approaches 4, substitute 4 into the function to give \(L = 4+2 = 6\).

Step 2: Setting Up the \(\varepsilon-\delta\) Definition

We want to show that for any positive number \(\varepsilon\), there always exists a positive number \(\delta\) such that if \(0 < |x - 4|< \delta\), then \(|f(x) - 6|< \varepsilon\). Now, \(|f(x) - 6| = |x+2 - 6| = |x - 4|\).

Step 3: Applying the \(\varepsilon-\delta\) Definition

To make \(|x - 4|< \varepsilon\), you can set \(\delta\) to be equal to \(\varepsilon\). So, if \(0 < |x - 4|< \varepsilon\), then \(|f(x) - 6| = |x - 4|< \varepsilon\), which conforms with the \(\varepsilon-\delta\) definition of limit.

Step 4: Conclusion

So, we have proved that \(L = 6\) is indeed the limit of \(f(x) = x+2\) as \(x\) approaches 4 using the \(\varepsilon-\delta\) definition of limit.

Key Concepts

Limit of a function

In calculus, one of the foundational concepts is the "limit of a function". Limits allow us to explore the behavior of a function as the input approaches a certain point. This is crucial for understanding and analyzing the continuity and differentiability of functions. In our particular problem, we focused on the limit of the function \( f(x) = x + 2 \) as \( x \) approaches 4.

To find this limit, we substitute the value of \( x \) into the function. For our case, substituting 4 into \( f(x) = x + 2 \) gives \( f(4) = 4 + 2 = 6 \). Hence, the limit is 6. This substitution process provides quick results, especially for simple functions. However, the formal way to verify this result involves the epsilon-delta definition, which solidifies the idea that the limit truly is 6.

Continuous functions

A function is called continuous at a point if small changes in the input around that point result in small changes in the output. More formally, a function \( f \) is continuous at a point \( c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \). This means there shouldn’t be any abrupt jumps, breaks, or holes at that point.

In our example, the function \( f(x) = x + 2 \) is continuous because when \( x \) approaches 4, not only does the limit exist, but the value of the function at that point, \( f(4) = 6 \), matches the limit we found. Continuous functions typically have a graph that can be drawn without lifting the pencil from the paper. Functions like linear, quadratic, and many polynomial functions are inherently continuous.

Calculus proof techniques

Calculus often requires proof methods that provide formal guarantees about properties like limits and continuity. One of these methods is the \( \varepsilon - \delta \) definition of limit, which we utilized in our exercise to prove the limit of the function \( f(x) = x + 2 \) as \( x \) approaches 4.

According to this definition, for every positive number \( \varepsilon \) (however small), there must exist a positive number \( \delta \) such that whenever \( 0 < |x - 4| < \delta \), it follows that \( |f(x) - 6| < \varepsilon \). In this case, after rearranging, we found \( |f(x) - 6| = |x - 4| \).

This simplification, \( |x - 4| < \varepsilon \), implies that we can choose \( \delta = \varepsilon \). Thus, the verification is straightforward because both sides simplify identically. This epsilon-delta approach might seem intricate, but it is a powerful technique that underpins rigorous mathematical proofs in calculus.