Proving The Limit Of A Quotient Lim (n→∞) X_n/y_n = X/y

Introduction

In the realm of mathematical analysis, understanding the behavior of sequences and their limits is paramount. One fundamental concept involves the limit of a quotient of two sequences. Specifically, we aim to rigorously prove that if a sequence $x_n$ converges to a limit $x$ and another sequence $y_n$ converges to a limit $y$, where $y$ is non-zero, then the sequence formed by the quotients $\frac{x_n}{y_n}$ converges to the quotient of the limits $\frac{x}{y}$. This seemingly intuitive result requires careful justification using the formal definitions of limits and convergence. In this comprehensive exploration, we will dissect the theorem, provide a detailed proof, and discuss its implications and applications.

Prerequisites: Limits and Convergence

Before diving into the proof, let's establish the foundational concepts of limits and convergence. A sequence $(x_n)$ is said to converge to a limit $x$ if, for every positive real number $\epsilon$, there exists a positive integer $N$ such that for all $n > N$, the absolute difference between $x_n$ and $x$ is less than $\epsilon$. Mathematically, this is expressed as:

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

This definition essentially states that as $n$ becomes sufficiently large, the terms of the sequence get arbitrarily close to the limit $x$. Similarly, a sequence $(y_n)$ converges to a limit $y$ if:

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

The condition $y \\neq 0$ is crucial. Division by zero is undefined, and the theorem would not hold if $y$ were zero. We also need to consider the implications of $y_n$ approaching zero, which we will address later in the proof.

Theorem Statement

Theorem: Let $(x_n)$ and $(y_n)$ be sequences of real numbers such that $\\lim_{n \\rightarrow \\infty} x_n = x$ and $\\lim_{n \\rightarrow \\infty} y_n = y$, where $y \\neq 0$. Then,

$$\\\\lim_{n \\rightarrow \\infty} \\frac{x_n}{y_n} = \\frac{x}{y}$$

This theorem is a cornerstone of limit theory, allowing us to compute limits of more complex expressions by breaking them down into simpler components. It is widely used in calculus, real analysis, and other areas of mathematics.

Proof

To prove the theorem, we need to show that for any given $\\epsilon > 0$, there exists an integer $N$ such that for all $n > N$, the absolute difference between $\\frac{x_n}{y_n}$ and $\\frac{x}{y}$ is less than $\\epsilon$. That is, we need to show:

$$\\\\forall \\epsilon > 0, \\\\exists N \\\\in \\\\mathbb{N} \\\\text{ such that } \\\\forall n > N, |\\frac{x_n}{y_n} - \\frac{x}{y}| < \\epsilon$$

We will proceed with the proof in several steps to make it more understandable.

Step 1: Manipulating the Expression

First, we manipulate the expression inside the absolute value to make it easier to work with:

$$\\\\|\\frac{x_n}{y_n} - \\frac{x}{y}| = |\\frac{x_n y - x y_n}{y_n y}| = |\\frac{x_n y - x y + x y - x y_n}{y_n y}| = |\\frac{(x_n - x)y - x(y_n - y)}{y_n y}|$$

Using the triangle inequality, we can further break down the expression:

$$\\\\|\\frac{(x_n - x)y - x(y_n - y)}{y_n y}| \\\\leq \\\\frac{|(x_n - x)y| + |x(y_n - y)|}{|y_n y|} = \\\\frac{|x_n - x||y| + |x||y_n - y|}{|y_n||y|}$$

Step 2: Bounding $\\frac{1}{|y_n|}$

Since $y \\neq 0$ and $y_n$ converges to $y$, we need to ensure that $y_n$ is not too close to zero for sufficiently large $n$. We can find an $N_1$ such that for all $n > N_1$, $|y_n - y| < \\frac{|y|}{2}$. This implies:

$$\\\|y| = |y_n - (y_n - y)| \\\\leq |y_n| + |y_n - y| < |y_n| + \\frac{|y|}{2}$$

Subtracting $\\frac{|y|}{2}$ from both sides, we get:

\\\\rac{|y|}{2} < |y_n|

Taking the reciprocal, we have:

$$\\\\frac{1}{|y_n|} < \\\\frac{2}{|y|}$$

This inequality provides a crucial bound for $\\frac{1}{|y_n|}$ for $n > N_1$.

Step 3: Choosing N and Applying Convergence Definitions

Now, we return to our inequality:

$$\\\\frac{|x_n - x||y| + |x||y_n - y|}{|y_n||y|} < \\\\frac{|x_n - x||y| + |x||y_n - y|}{\\frac{|y|}{2}|y|} = \\\\frac{2}{|y|^2}(|x_n - x||y| + |x||y_n - y|)$$

Since $\\lim_{n \\rightarrow \\infty} x_n = x$, for any $\\epsilon_1 > 0$, there exists an $N_2$ such that for all $n > N_2$, $|x_n - x| < \\epsilon_1$. Similarly, since $\\lim_{n \\rightarrow \\infty} y_n = y$, for any $\\epsilon_2 > 0$, there exists an $N_3$ such that for all $n > N_3$, $|y_n - y| < \\epsilon_2$.

Let's choose $\\epsilon_1 = \\\\frac{\\epsilon |y|^2}{4|y|}$ and $\\epsilon_2 = \\\\frac{\\epsilon |y|^2}{4|x|}$ (assuming $x \\neq 0$; if $x = 0$, we can modify the argument slightly). Then, for $n > N_2$, we have $|x_n - x| < \\\\frac{\\epsilon |y|^2}{4|y|}$, and for $n > N_3$, we have $|y_n - y| < \\\\frac{\\epsilon |y|^2}{4|x|}$.

Now, let $N = \\max(N_1, N_2, N_3)$. For all $n > N$, we have:

$$\\\\|\\frac{x_n}{y_n} - \\frac{x}{y}| < \\\\frac{2}{|y|^2}(|x_n - x||y| + |x||y_n - y|) < \\\\frac{2}{|y|^2}(\\\\frac{\\epsilon |y|^2}{4|y|}|y| + |x|\\\\frac{\\epsilon |y|^2}{4|x|}) = \\\\frac{2}{|y|^2}(\\\\frac{\\epsilon |y|^2}{4} + \\\\frac{\\epsilon |y|^2}{4}) = \\epsilon$$

Thus, we have shown that for any $\\epsilon > 0$, there exists an $N$ such that for all $n > N$, $|\\frac{x_n}{y_n} - \\frac{x}{y}| < \\epsilon$. This completes the proof.

Step 4: Handling the Case When x = 0

If x = 0, then we want to show that $\\lim_{n \\rightarrow \\infty} \\frac{x_n}{y_n} = 0$. We have:

$$\\\\|\\frac{x_n}{y_n}| = |x_n| \\\\cdot \\\\frac{1}{|y_n|} < |x_n| \\\\cdot \\\\frac{2}{|y|}$$

Since $\\lim_{n \\rightarrow \\infty} x_n = 0$, for any $\\epsilon > 0$, there exists an $N_2$ such that for all $n > N_2$, $|x_n| < \\\\frac{\\epsilon |y|}{2}$. Let $N = \\max(N_1, N_2)$. Then, for all $n > N$:

$$\\\\|\\frac{x_n}{y_n}| < \\\\frac{\\epsilon |y|}{2} \\\\cdot \\\\frac{2}{|y|} = \\epsilon$$

This shows that $\\lim_{n \\rightarrow \\infty} \\frac{x_n}{y_n} = 0$ when $x = 0$.

Implications and Applications

The theorem we have proven has significant implications in various areas of mathematics. It allows us to compute limits of rational functions, which are functions formed by the ratio of two polynomials. For example, consider the limit:

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

We can divide both the numerator and the denominator by $n^2$:

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

Using the theorem, we can find the limit of the numerator and the denominator separately:

$$\\\\lim_{n \\rightarrow \\infty} (3 + \\frac{2}{n} + \\frac{1}{n^2}) = 3$$

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

Thus, the limit of the quotient is $\\frac{3}{2}$.

This theorem is also crucial in the study of continuity and differentiability of functions. It provides a rigorous foundation for many calculus techniques and is indispensable for advanced mathematical analysis.

Conclusion

We have provided a detailed and rigorous proof of the theorem stating that the limit of a quotient of two convergent sequences is equal to the quotient of their limits, provided the limit of the denominator is non-zero. This theorem is a fundamental result in real analysis and has wide-ranging applications in calculus and beyond. Understanding this theorem and its proof is essential for anyone seeking a deeper understanding of mathematical analysis. The step-by-step approach we have taken, including manipulating the expression, bounding terms, and carefully applying the definitions of convergence, highlights the intricacies and elegance of mathematical reasoning.