Prove Limit Of (2n-1)/(3n+2) As N Approaches Infinity

In the realm of mathematical analysis, understanding limits is fundamental. Limits form the bedrock upon which concepts like continuity, derivatives, and integrals are built. Among the various types of limits, the limit of a sequence is a crucial starting point. This article delves into the formal proof of a specific limit: ${\lim_{n \rightarrow \infty} \frac{2n-1}{3n+2} = \frac{2}{3}}$. We will dissect the problem, explore the epsilon-delta definition of a limit, and construct a rigorous proof that elucidates why this limit holds true. This exploration will not only enhance your understanding of limit proofs but also provide a template for tackling similar problems in mathematical analysis. Mastering these techniques is essential for anyone delving deeper into calculus and real analysis, as they provide the tools necessary to rigorously define and manipulate the core concepts of these fields.

Formal Definition of a Limit

Before we embark on the proof, it's imperative to understand the formal definition of a limit of a sequence. This definition provides the rigorous framework we need to construct a sound argument. We say that the limit of a sequence ${(a_n)}$ as ${n}$ approaches infinity is ${L}$, written as ${\lim_{n \rightarrow \infty} a_n = L}$, if for every real number ${\epsilon > 0}$, there exists a natural number ${N}$ such that for all ${n > N}$, the absolute difference between ${a_n}$ and ${L}$ is less than ${\epsilon}$. In mathematical notation, this is expressed as:

$$\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ such that } \forall n > N, |a_n - L| < \epsilon$$

This definition, while seemingly complex at first glance, encapsulates the intuitive notion of a limit. It states that we can make the terms of the sequence arbitrarily close to the limit ${L}$ by going far enough out in the sequence (i.e., by choosing a sufficiently large ${n}$). The ${\epsilon}$ represents how close we want the terms to be to the limit, and the ${N}$ tells us how far out in the sequence we need to go to achieve that closeness. The essence of proving a limit lies in finding such an ${N}$ for any given ${\epsilon}$. Understanding this definition is crucial as it forms the basis for all limit proofs and is a cornerstone of real analysis. The ability to interpret and apply this definition is a key skill for any mathematician or student of mathematics.

Problem Statement and Initial Analysis

Our specific problem is to prove that the limit of the sequence ${a_n = \frac{2n-1}{3n+2}}$ as ${n}$ approaches infinity is ${\frac{2}{3}}$. That is, we need to show:

$$\lim_{n \rightarrow \infty} \frac{2n-1}{3n+2} = \frac{2}{3}$$

To tackle this, we need to demonstrate that for any arbitrarily small positive number ${\epsilon}$, we can find a natural number ${N}$ such that for all ${n > N}$, the inequality ${|\frac{2n-1}{3n+2} - \frac{2}{3}| < \epsilon}$ holds. The initial step in any limit proof is to manipulate the expression inside the absolute value to make it easier to work with. This often involves finding a common denominator, simplifying the expression, and attempting to isolate ${n}$. In this case, our goal is to find an expression of the form ${\frac{C}{f(n)}}$, where ${C}$ is a constant and ${f(n)}$ is a function of ${n}$, as this will allow us to relate the expression to ${\epsilon}$ and ultimately solve for ${N}$. This initial analysis is critical because it lays the groundwork for the formal proof and provides a roadmap for the subsequent steps. It requires algebraic manipulation skills and a good understanding of inequalities.

Detailed Proof

Let's begin by manipulating the expression inside the absolute value:

$$\left| \frac{2n-1}{3n+2} - \frac{2}{3} \right| = \left| \frac{3(2n-1) - 2(3n+2)}{3(3n+2)} \right|$$

Simplifying the numerator, we get:

$$\left| \frac{6n - 3 - 6n - 4}{9n + 6} \right| = \left| \frac{-7}{9n + 6} \right|$$

Since ${n}$ is a natural number, ${9n + 6}$ is always positive. Therefore, we can remove the absolute value from the denominator:

$$\left| \frac{-7}{9n + 6} \right| = \frac{7}{9n + 6}$$

Now, we want to find an ${N}$ such that for all ${n > N}$, ${\frac{7}{9n + 6} < \epsilon}$. To do this, we solve the inequality for ${n}$:

$$\frac{7}{9n + 6} < \epsilon$$

Multiplying both sides by ${9n + 6}$ and dividing by ${\epsilon}$ (since ${\epsilon > 0}$), we get:

$$\frac{7}{\epsilon} < 9n + 6$$

Subtracting 6 from both sides:

$$\frac{7}{\epsilon} - 6 < 9n$$

Dividing by 9:

$$\frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right) < n$$

Now, we can choose ${N}$ to be any integer greater than ${\frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right)}$. To ensure ${N}$ is a natural number, we can take the smallest integer greater than this value by using the ceiling function:

$$N = \left\lceil \frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right) \right\rceil$$

This choice of ${N}$ guarantees that for all ${n > N}$, the inequality ${|\frac{2n-1}{3n+2} - \frac{2}{3}| < \epsilon}$ holds. This step-by-step derivation is essential for understanding the logical flow of the proof and how each step contributes to the final result. It highlights the importance of algebraic manipulation, solving inequalities, and choosing an appropriate ${N}$.

Formal Proof Summary

To formally prove that ${\lim_{n \rightarrow \infty} \frac{2n-1}{3n+2} = \frac{2}{3}}$, we follow these steps:

1. State the goal: We want to show that for every ${\epsilon > 0}$, there exists an ${N \in \mathbb{N}}$ such that for all ${n > N}$, ${|\frac{2n-1}{3n+2} - \frac{2}{3}| < \epsilon}$.

2. Manipulate the expression: We simplified ${|\frac{2n-1}{3n+2} - \frac{2}{3}|}$ to ${\frac{7}{9n + 6}}$.

3. Find N: We solved the inequality ${\frac{7}{9n + 6} < \epsilon}$ for ${n}$ and found that ${n > \frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right)}$. We then chose ${N = \left\lceil \frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right) \right\rceil}$.

4. Write the formal proof:

   Let ${\epsilon > 0}$ be given. Choose ${N = \left\lceil \frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right) \right\rceil}$. Then for all ${n > N}$, we have:

   $$\left| \frac{2n-1}{3n+2} - \frac{2}{3} \right| = \frac{7}{9n + 6} < \frac{7}{9N + 6} \leq \frac{7}{9 \left( \frac{1}{9} \left( \frac{7}{\epsilon} - 6 \right) \right) + 6} = \epsilon$$

   Therefore, by the definition of a limit, ${\lim_{n \rightarrow \infty} \frac{2n-1}{3n+2} = \frac{2}{3}}$. This formal summary is essential because it encapsulates the entire proof in a concise and logical manner. It demonstrates the application of the epsilon-delta definition and showcases the key steps involved in proving a limit. The ability to present a proof in this formal style is a crucial skill in mathematical analysis.

Common Mistakes and How to Avoid Them

Proving limits can be challenging, and there are several common mistakes students often make. One frequent error is not correctly manipulating the expression inside the absolute value. For instance, failing to find a common denominator or making algebraic errors while simplifying can lead to an incorrect result. It's crucial to double-check each step of the algebraic manipulation to avoid these errors. Another common mistake is choosing an inappropriate ${N}$. The chosen ${N}$ must guarantee that the inequality ${|a_n - L| < \epsilon}$ holds for all ${n > N}$. A common error is to find a value of ${n}$ that satisfies the inequality for a specific ${\epsilon}$ but not for all ${n > N}$. To avoid this, always solve the inequality for ${n}$ and choose ${N}$ to be an integer greater than the resulting expression. Another pitfall is not stating the formal definition of a limit clearly at the beginning of the proof. Explicitly stating the definition helps to guide the proof and ensures that all necessary steps are included. Finally, some students struggle with the logical flow of the proof. It's important to clearly articulate each step and explain why it is valid. This includes justifying the choice of ${N}$ and demonstrating that the inequality holds for all ${n > N}$. By being mindful of these common mistakes and practicing limit proofs, you can develop a strong understanding of this fundamental concept in mathematical analysis.

Conclusion

In this article, we have rigorously proven that ${\lim_{n \rightarrow \infty} \frac{2n-1}{3n+2} = \frac{2}{3}}$ using the epsilon-delta definition of a limit. We dissected the problem, performed the necessary algebraic manipulations, and carefully chose ${N}$ to ensure the inequality held for all ${n > N}$. This process highlights the importance of understanding the formal definition of a limit and mastering the techniques of algebraic manipulation and inequality solving. The ability to prove limits is a fundamental skill in mathematical analysis, and it lays the groundwork for understanding more advanced concepts such as continuity, derivatives, and integrals. By understanding the step-by-step approach outlined in this article, you can confidently tackle similar limit problems and deepen your understanding of mathematical analysis. The key takeaway is that proving limits requires a combination of algebraic skill, a thorough understanding of the definition, and a logical approach to constructing a sound argument. This exploration of limit proofs not only enhances your mathematical toolkit but also cultivates critical thinking and problem-solving skills that are valuable in various fields.