Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give an \(\varepsilon-\delta\) proof of each limit fact. $$ \lim _{x \rightarrow 0}\left(\frac{2 x^{2}-x}{x}\right)=-1 $$

Step-by-Step Solution

Answer

The limit \( \lim_{x \to 0} \frac{2x^2 - x}{x} = -1 \) is proven using \( \varepsilon-\delta \) criteria.

Step 1: Simplify the Function

First, let's simplify the given function \( f(x) = \frac{2x^2 - x}{x} \). We can separate the terms:\[\frac{2x^2 - x}{x} = \frac{2x^2}{x} - \frac{x}{x} = 2x - 1\]This simplifies the function to \( f(x) = 2x - 1 \).

Step 2: Identify the Limit to Prove

Since we want to show that \( \lim_{x \to 0} (2x - 1) = -1 \), our task is to prove that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that \[|f(x) + 1| < \varepsilon \quad \text{whenever} \quad 0 < |x| < \delta\]This translates to proving:\[ |2x - 1 + 1| = |2x| < \varepsilon\].

Step 3: Find a Suitable Delta

To ensure \( |2x| < \varepsilon \), observe that:\[|2x| = 2|x| < \varepsilon\]Dividing both sides of the inequality by 2 gives:\[|x| < \frac{\varepsilon}{2}\]Therefore, we choose \( \delta = \frac{\varepsilon}{2} \).

Step 4: Provide the Epsilon-Delta Proof

For \( \varepsilon > 0 \), choose \( \delta = \frac{\varepsilon}{2} \). Then, whenever \( 0 < |x| < \delta \), we have:\[|2x| = 2|x| < 2\delta = 2\left(\frac{\varepsilon}{2}\right) = \varepsilon\]Thus, \( |2x| < \varepsilon \), which implies \( |2x - 1 + 1| < \varepsilon \), and therefore, \( |f(x) + 1| < \varepsilon \).

Step 5: Conclusion: Limit is Proven

Since we've shown that for every \( \varepsilon > 0 \), there exists a \( \delta = \frac{\varepsilon}{2} \) such that whichever positive \( |x| < \delta \), \(|2x - 1 + 1| < \varepsilon\), we conclude that:\[ \lim_{x \to 0} \frac{2x^2 - x}{x} = -1\] as required by the limit definition.

Key Concepts

Limit Calculus

In calculus, limits are used to describe the behavior of functions as they approach a particular point, often leading to fundamental ideas such as continuity and derivatives. The limit of a function at a given point tells us what value the function approaches as the input approaches that point.
For example, in the exercise given, we are interested in finding the limit of the function \( f(x) = \frac{2x^2 - x}{x} \) as \( x \to 0 \). By simplifying \( f(x) \) to \( 2x - 1 \), it is apparent that as \( x \) approaches 0, the function heads towards -1.
This calculation is central in many areas of calculus because it helps in understanding the subtle behavior of functions around points that cause complications, such as zero denominators or undefined expressions. Limits are foundational for derivatives and integrals, which in turn are vital for the broader analysis of functions.

Limits and Continuity

Limits are closely tied to the concept of continuity, which refers to a function's behavior when there are no sudden changes or breaks. A function is considered continuous at a point when the limit at that point exists and equals the function's value there.
In our original exercise, simplifying the function showed us that it can be continuous as \( x \to 0 \), even though the division by zero is initially a concern. Calculus often uses limits to 'smooth over' these concerns, effectively allowing functions to maintain continuity through analytical methods.
This is beneficial because continuity assures that a function behaves predictably, and helps mathematicians and students understand real-world phenomena where slight changes lead to small, predictable outcomes. It plays a crucial role in various fields such as physics, engineering, and economics, where understanding continuous changes are essential.

Mathematical Proofs

Mathematical proofs are rigorous arguments that use logic to demonstrate the truth of a statement. In calculus, proof techniques like the epsilon-delta method provide a formal foundation for understanding limits.
The epsilon-delta proof is a mathematical technique where, for any arbitrarily small positive number \( \varepsilon \), we find a \( \delta \) such that the function remains within \( \varepsilon \) of the limit whenever the variable is within \( \delta \) of the point.
This method is used in the exercise to formally prove that \( \lim_{x \to 0} (2x - 1) = -1 \). By choosing \( \delta = \frac{\varepsilon}{2} \), we ensure that whenever \( |x| < \delta \), the values of \( 2x \) remain bounded by \( \varepsilon \), thus establishing the required proof.
The rigor of mathematical proofs like epsilon-delta ensures that mathematical statements are universally valid, and they form the bedrock of theoretical understanding and application of calculus across all scientific disciplines.