Problem 21 - Calculus: Early Transcendentals

Question

Limit proofs Use the precise definition of a limit to prove the following limits. Specify a relationship between \(\varepsilon\) and \(\delta\) that guarantees the limit exists. $$\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}=8(\text { Hint: Factor and simplify. })$$

Step-by-Step Solution

Answer

Question: Prove the limit $\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}=8$ using the precise definition of a limit (\(\varepsilon-\delta\) definition) and find the relationship between \(\varepsilon\) and \(\delta\) that guarantees the limit exists. Answer: Based on the \(\varepsilon-\delta\) definition and the steps we have followed, we have found the relationship between \(\varepsilon\) and \(\delta\): \(\delta = \varepsilon\). This relationship ensures the existence of the given limit as per the precise definition of a limit.

Step 1: Simplify the function \(f(x)\)

We are given the function \(f(x) = \frac{x^2 - 16}{x - 4}\). To simplify this, we can factor the numerator: $$x^2 - 16 = (x + 4)(x - 4)$$ Now, we can simplify the fraction by canceling out the \((x - 4)\) terms: $$f(x) = \frac{(x + 4)(x - 4)}{x - 4} = x + 4$$

Step 2: Find an expression for \(|f(x) - L|\) in terms of \(x\)

Now that we have our simplified function \(f(x) = x + 4\), let's find the expression for \(|f(x) - L|\): $$|f(x) - L| = |(x + 4) - 8| = |x - 4|$$

Step 3: Derive a relationship between \(\varepsilon\) and \(\delta\)

We want to find a relationship between \(\varepsilon\) and \(\delta\) such that if \(0 < |x - 4| < \delta\), then \(|x - 4| < \varepsilon\). Since both inequalities involve the same expression, we can choose \(\delta = \varepsilon\). For any given \(\varepsilon > 0\), if we choose \(\delta = \varepsilon\) and have \(0 < |x - 4| < \delta\), then it follows that \(|f(x) - L| = |x - 4| < \varepsilon\). This proves the limit using the precise definition and establishes a relationship between \(\varepsilon\) and \(\delta\): \(\delta = \varepsilon\).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is the formal way to prove the existence of a limit. In simpler terms, it tells us how close we need to get to a certain point to make a function's value close to a specific number. Usually, we use this method to prove that \( \lim_{x \to a} f(x) = L \). The key players in this method are \( \varepsilon \) (epsilon) and \( \delta \) (delta).

• \( \varepsilon \) represents how close the function \( f(x) \) should be to the limit \( L \).
• \( \delta \) corresponds to how close \( x \) needs to be to \( a \) to ensure the function's proximity to \( L \).

To establish a limit, we need to choose \( \delta \) such that whenever \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \varepsilon \). Essentially, small \( \delta \) values prove that \( f(x) \) gets closer to \( L \) as \( x \) approaches \( a \). This conceptual framework ensures a function's behavior as \( x \) approaches a particular point, seamlessly matching our expectations of limits in calculus.

Limit Simplification

Simplifying a limit is often the first step in solving it, especially when dealing with expressions that initially appear complex or undefined. For the function \( f(x) = \frac{x^2 - 16}{x - 4} \), it appears undefined at \( x = 4 \) due to division by zero.To simplify, factor the numerator:

• \( x^2 - 16 = (x + 4)(x - 4) \)

This allows us to cancel the \( (x - 4) \) terms, leading to \( f(x) = x + 4 \) when \( x eq 4 \). With this simplification, it becomes clearer to analyze and evaluate the limit. Simplifying limits is crucial because it helps identify the function’s true behavior around the value that \( x \) approaches. Once the simplification reveals \( f(x) = x + 4 \), finding the limit as \( x \to 4 \) becomes trivial, as it is just \( 8 \). Simplification not only gives insight but also makes subsequent calculations straightforward.

Rational Expressions

Rational expressions are fractions where the numerator and/or the denominator are polynomials. Understanding these expressions is key when dealing with limits, as they can often result in indeterminate forms like \( \frac{0}{0} \). For the expression \( \frac{x^2 - 16}{x - 4} \), recognizing it as a rational expression directs us to simplify by factoring.Usually, simplifying rational expressions involves:

• Identifying common factors in the numerator and denominator.
• Canceling common factors to avoid indeterminate forms.

By simplifying, \( f(x) = \frac{(x+4)(x-4)}{x-4} \) becomes \( f(x) = x+4 \), providing a continuous function wherever \( x eq 4 \). Rational expressions often require careful study to avoid incorrect cancellation and to ensure limits are calculated correctly. Mastery of rational expressions helps avoid pitfalls when handling limits in calculus.

Factoring Polynomials

Factoring polynomials is a fundamental process in algebra and calculus, crucial for simplifying expressions and solving equations. When facing complex expressions like \( x^2 - 16 \), factoring is an essential step towards simplification.In this context, \( x^2 - 16 \) is a difference of squares, which factors neatly into:

• \( x^2 - 16 = (x+4)(x-4) \)

Recognizing and applying patterns like the difference of squares can greatly simplify algebraic operations. It also ensures accurate cancellation in expressions like \( \frac{(x+4)(x-4)}{x-4} \), allowing us to redefine the function for limit calculations. Factoring helps avoid errors and streamline the process of determining precise values in calculus problems. Having a strong grasp of polynomial factoring grants confidence in breaking down and understanding complex algebraic expressions, making calculus problems much more approachable.