Problem 19 - Calculus Early Transcendentals

Question

Use the precise definition of a limit to prove the following limits. $$\lim _{x \rightarrow 1}(8 x+5)=13$$

Step-by-Step Solution

Answer

Question: Prove the following limit using the precise definition of a limit: $$\lim_{x \rightarrow 1}(8x + 5) = 13$$ Answer: By using the precise definition of a limit, we showed that for every ε > 0, there exists a δ > 0 (specifically, δ = ε/8) such that if 0 < |x - 1| < δ, then |(8x + 5) - 13| < ε. This confirms that the given limit is true: $$\lim_{x \rightarrow 1}(8x + 5) = 13$$

Step 1: Write down the definition of a limit

According to the precise definition of a limit, a function f(x) has a limit L as x approaches c, written as: $$\lim_{x \rightarrow c}f(x) = L$$ if for every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε.

Step 2: Substitute the given function and limit into the definition

In this case, we have the function f(x) = 8x + 5, the point c = 1, and the limit L = 13. So, we'll substitute these values into the definition: For every ε > 0, there exists a δ > 0 such that if 0 < |x - 1| < δ, then |(8x + 5) - 13| < ε.

Step 3: Show that there exists a δ > 0 for every ε > 0

Let's find the value of δ that satisfies the condition. First, simplify the expression inside the absolute value |(8x + 5) - 13|: |(8x + 5) - 13| = |8x - 8| = 8|x - 1|. Now, we want to find a δ such that 8|x - 1| < ε when 0 < |x - 1| < δ. Let's choose δ = ε/8. By doing this, we can show that 0 < |x - 1| < δ implies 8|x - 1| < ε: Suppose 0 < |x - 1| < δ = ε/8 Then 8|x - 1| < 8(ε/8) = ε. By choosing δ = ε/8, we have shown that there exists a δ > 0 such that for every ε > 0, the inequality 0 < |x - 1| < δ implies |(8x + 5) -13| < ε. Therefore, according to the precise definition of a limit, we have proved: $$\lim _{x \rightarrow 1}(8 x+5)=13$$

Key Concepts

Limit Proofs

A limit proof is a method used to demonstrate that a function approaches a specific value as the input approaches a certain point. To write a limit proof, one uses the precise definition of a limit, which involves setting conditions using epsilon (\( \varepsilon \)) and delta (\( \delta \)). In this kind of proof:

• We begin by stating what we want to prove. For instance, proving that \( \lim_{x \rightarrow 1}(8x+5)=13 \).
• Next, we employ the epsilon-delta definition, which ensures that the function value \( f(x) \) comes arbitrarily close to the limit \( L \) by choosing a suitable \( \delta \) for any given \( \varepsilon \).
• The goal is to find a corresponding \( \delta \) such that if \( 0 < |x - c| < \delta \), then \( |f(x) - L| < \varepsilon \).

Practically, this involves manipulating inequalities and understanding the behavior of the function. Conclusively, the choice of \( \delta \) completes the proof, verifying the limit of a function.

Epsilon-Delta Definition

The epsilon-delta definition is essential in understanding the concept of limits in calculus. This definition formalizes what it means for a function to approach a limit as the input gets closer to a specified point. To apply it, follow these steps:

• Start with any \( \varepsilon > 0 \). This \( \varepsilon \) represents an arbitrary closeness to the limit \( L \).
• Find a corresponding \( \delta > 0 \) so that whenever \( 0 < |x - c| < \delta \), we have \( |f(x) - L| < \varepsilon \).

In simpler terms, \( \varepsilon \) defines how close \( f(x) \) should be to \( L \), while \( \delta \) gives a range around \( c \) where this closeness condition will hold. Using these parameters efficiently helps you logically show that the function almost touches the limit as \( x \) approaches \( c \). In the problem given, for every \( \varepsilon \), \( \delta = \varepsilon/8 \) ensures the condition \( 8|x - 1| < \varepsilon \) is satisfied, successfully applying the epsilon-delta definition.

Calculus Concepts

In calculus, the concept of limits is fundamental, as it is the foundational idea for derivatives and integrals. Understanding limits through precise definitions enriches the comprehension of how functions behave near specific points. Key concepts include:

• Continuity: A function is continuous at a point if the limit equals the function's value there.
• Derivatives: Derivatives are defined as the limit of the difference quotient as the interval approaches zero.
• Integrals: Integration is fundamentally connected to limits, specifically in defining the area under a curve via limits of Riemann sums.

Grasping these calculus concepts helps students solve a variety of mathematical problems, and using the epsilon-delta approach provides a rigorous method to work with limits. Practicing with different functions and limit proofs strengthens problem-solving skills and deepens understanding of calculus at large.