Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give an \(\varepsilon-\delta\) proof of each limit fact. $$ \lim _{x \rightarrow 1} \frac{14 x^{2}-20 x+6}{x-1}=8 $$

Step-by-Step Solution

Answer

The limit is correctly verified by \( \varepsilon-\delta \) with \( \delta = \frac{\varepsilon}{14} \).

Step 1: Understand the Problem

We want to prove that as \( x \) approaches 1, the function \( \frac{14x^2 - 20x + 6}{x - 1} \) approaches the value 8. This is expressed as \( \lim_{x \to 1} \frac{14x^2 - 20x + 6}{x - 1} = 8 \). We need to show the limit using the \( \varepsilon-\delta \) definition.

Step 2: Simplify the Expression

Notice that the given expression \( \frac{14x^2 - 20x + 6}{x - 1} \) appears to be indeterminate when \( x = 1 \). Let's try to simplify it. The polynomial in the numerator can be factored:\(14x^2 - 20x + 6 = (2x - 2)(7x - 3).\)So the expression becomes:\(\frac{(2x - 2)(7x - 3)}{x - 1} = (2)(7x - 3) = 14x - 6.\)

Step 3: Rewrite the Limit

Now, the limit can be more easily handled:\(\lim_{x \to 1} (14x - 6).\)Substituting \( x = 1 \) gives:\(14(1) - 6 = 8.\)Thus, \( \lim_{x \to 1} \frac{14x^2 - 20x + 6}{x - 1} = 8 \) is plausible after simplification.

Step 4: Setup the \(\varepsilon-\delta\) Definition

To prove this using \( \varepsilon-\delta \), we need to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 1| < \delta \), then \( |(14x - 6) - 8| < \varepsilon \).

Step 5: Calculate the \(\varepsilon-\delta\) Condition

We need \( |14x - 14| < \varepsilon \) since:\(|(14x - 6) - 8| = |14x - 14|.\)This simplifies to:\(|14(x - 1)| < \varepsilon \Rightarrow |x - 1| < \frac{\varepsilon}{14}.\)

Step 6: Choose \(\delta\) Based on \(\varepsilon\)

From step 5, we choose \( \delta = \frac{\varepsilon}{14} \). Therefore, for every \( \varepsilon > 0 \), by choosing \( \delta = \frac{\varepsilon}{14} \), the condition of the \( \varepsilon-\delta \) definition is satisfied.

Step 7: Conclude the Proof

Thus, we've shown that for every \( \varepsilon > 0 \), we can find a \( \delta = \frac{\varepsilon}{14} \) ensuring \( |x - 1| < \delta \) implies \( |(14x - 6) - 8| < \varepsilon \). This confirms that \( \lim_{x \to 1} \frac{14x^2 - 20x + 6}{x - 1} = 8 \) by the \( \varepsilon-\delta \) definition.

Key Concepts

epsilon-delta proof

The epsilon-delta definition is a rigorous way to confirm a limit statement in calculus. It formalizes the intuitive notion of what it means for a function to "approach" a certain value. To understand an epsilon-delta ( \(\varepsilon-\delta\)) proof, remember:

• For any positive value of \(\varepsilon\), representing how close we want the function's output to be to the limit, there should exist a corresponding positive value of \(\delta\), indicating how close \(x\) should be to the target point, such that:
• If \(|x - c| < \delta\), then \(|f(x) - L| < \varepsilon\).

In simpler terms, no matter how small of a margin \(\varepsilon\) that we want between the function value and the limit \(L\), we can find a range, \(\delta\), around the number \(c\) where the input function will stay within that margin. This approach prevents any hypothetical "wiggle room" that could contradict the calculated limit. It solidifies the computation and confidence in the limit value you are seeking.

evaluating limits

Evaluating limits is crucial in determining the behavior of a function as it approaches a particular point. The process usually involves:

• Checking if the function is defined at the point.
• Having a preliminary check if substitution works without hitting indeterminacy like \(\frac{0}{0}\).
• Using algebraic manipulation or simplification in cases of indeterminate forms.

In our exercise, direct substitution into the function led to an indeterminate form, showing the importance of the proper analysis and simplification before evaluating the limit.
By factoring and canceling inevitable troublesome terms, we reduced complexity, revealing that what initially appeared complex was really quite simple once the problem was unraveled correctly.

polynomial simplification

Simplifying polynomials is often necessary when dealing with limits, especially when the substitution yields indeterminate forms. Here are simple steps:

• Factor the polynomial if possible. Identifying and extracting common factors simplifies expressions.
• Cancel zero-inducing terms to neutralize indeterminacy.
• Then apply the simplified expression in the limit context again.

For example, the polynomial \(14x^2 - 20x + 6\) factors into \((2x - 2)(7x - 3)\). Because it's being divided by \(x - 1\), we cancel the zero-producing term with an equivalent expression, resulting in an expression of \(14x - 6\) which evaluates easily.
This simplification turns a bothersome expression into a straightforward calculation, facilitating finding the limit accurately.

limit definition

The definition of a limit is foundational in calculus. It describes the value that a function approaches as the input approaches some value. Formally:

• \(\lim_{x \to c}f(x) = L\), indicating that \(f(x)\) gets arbitrarily close to \(L\) as \(x\) approaches \(c\).
• This notion serves as a bridge between algebra and calculus by formally introducing the behavior of functions near points of interest, even if those points are not included in the domain.
• Limits help define other calculus concepts like continuity, derivatives, and integrals.

In this exercise, the polynomial demanded limit attention due to an indeterminate specification \((\frac{0}{0})\). Hence, grasping the depth of limits involves understanding the algebraic manipulations necessary for simplification and corresponding proof strategies that ultimately demonstrate a function's behavior as it nears a specific point.