Problem 48 - 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 \(L .\) $$ \lim _{x \rightarrow-4}\left(x^{2}+4 x\right) $$

Step-by-Step Solution

Answer

The limit \(L\) of the function \(f(x) = x^2 + 4x\) as \(x\) approaches -4 is 0. By setting \(\delta = \varepsilon\), it's shown that the function satisfies the \(\varepsilon-\delta\) condition.

Step 1: Find the Limit

The limit of the function \(f(x)=x^{2}+4 x\) as \(x\) approaches -4 can be found by substituting -4 into the function: \(L = (-4)^{2}+4(-4) = 16 - 16 = 0\). So, the limit \(L\) is 0.

Step 2: Use the Epsilon-Delta Definition

To fit our function into the \(\varepsilon-\delta\) definition, let's start by expressing \(|f(x) - L|\) as \(|x^2 + 4x - 0|\). We can simplify this as \(|x(x+4)|\). Now for all \(\varepsilon > 0\), we need to find a \(\delta > 0\) such that if \(0 < |x - (-4)| = |x+4| < \delta\), then \(|f(x) - L| = |x(x + 4)| < \varepsilon\). Let's choose \(\delta = \varepsilon\). Note that \(|x+4| = \delta < \varepsilon\). So, this shows that for any positive \(\varepsilon\), we can choose \(\delta = \varepsilon\) and it will hold that if \(0 < |x + 4| < \delta\), then \( |x(x + 4)| < \varepsilon\). In other words, the limit of \(f(x)=x^{2}+4 x\) as \(x\) approaches -4 is 0, and this satisfies the \(\varepsilon-\delta\) definition of a limit.

Key Concepts

Limit of a Function

Understanding the concept of the limit of a function is crucial in calculus. It describes the behavior of a function as its input approaches a particular value. When we say the limit of a function \( f(x) \) as \( x \) approaches \( c \) is \( L \), we symbolize it by \( \lim_{x \to c} f(x) = L \). This suggests that as \( x \) gets closer to \( c \), \( f(x) \) gets closer to \( L \).
To find limits, we often substitute \( x \) with the approaching value. In the problem statement, substituting -4 into the given function \( f(x) = x^2 + 4x \) gives us 0. The limit is 0 as \( x \rightarrow -4 \). This calculation helps us verify the expected behavior of the function near that point.
Limits are pivotal because they lay the groundwork for more complex calculus concepts, like derivatives and integrals.

Limit Proof

Proving a limit using the \(\varepsilon-\delta\) definition provides reassurance of our conclusions. It's the formal statement that whenever \( x \) is close to a certain value, the function is near a certain point. In this case, we show that \( \lim_{x \to -4} (x^2 + 4x) = 0 \) using this definition.

• For a given \( \varepsilon > 0 \), we need to find \( \delta > 0 \) such that if \( 0 < |x + 4| < \delta \), then \( |x(x+4)| < \varepsilon \).
• Rewriting \( |f(x) - L| \) leads to \( |x(x+4)| \). We found that setting \( \delta = \varepsilon \) works here.

Hence, choosing a suitable \( \delta \) relative to \( \varepsilon \) fulfills the \(\varepsilon-\delta\) condition, confirming the limit is indeed 0.

Calculus Concept

The \(\varepsilon-\delta\) definition of limits is fundamental to understanding calculus. It provides a rigorous method to define what it means for a function to approach a certain value. It's one of the key calculus concepts that requires a firm grasp but also offers profound insights into how functions behave.

• \(\varepsilon\) represents how close the function's value needs to be to the limit.
• \(\delta\) signifies how close \( x \) must be to the approaching value.

This method is not about finding the limit but proving it, offering a solid foundation in calculus. As students, mastering this process prepares you for further studies in calculus, especially when dealing with more complex functions or in multivariable calculus. Understanding these concepts paves the way for exploring continuity, derivatives, and integrals in later calculus topics.