Problem 92 - Thomas Calculus

Question

In Exercises \(91-98\) , assume that each sequence converges and find its limit. $$ a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2} $$

Step-by-Step Solution

Answer

The limit of the sequence is 2.

Step 1: Understanding the Problem

We are given a sequence with the initial term \( a_1 = -1 \) and a recursive formula \( a_{n+1} = \frac{a_n + 6}{a_n + 2} \). Our task is to find the limit of this sequence as \( n \to \infty \).

Step 2: Assume the Limit Exists

Since the sequence is assumed to converge, let's denote its limit by \( L \). Thus, we assume that as \( n \to \infty \), \( a_n \to L \). Hence, \( a_{n+1} \to L \) too.

Step 3: Set Up the Limit Equation

By substituting \( L \) for both \( a_n \) and \( a_{n+1} \) in the recursive formula \( a_{n+1} = \frac{a_n + 6}{a_n + 2} \), we get the equation \( L = \frac{L+6}{L+2} \). This equation will help us to find \( L \).

Step 4: Solve the Limit Equation

Multiply both sides of the equation \( L = \frac{L+6}{L+2} \) by \( L + 2 \) to eliminate the fraction, which gives: \( L(L+2) = L + 6 \). Expanding the left side, we have \( L^2 + 2L = L + 6 \).

Step 5: Rearrange and Factor

Subtract \( L \) and 6 from both sides to obtain \( L^2 + 2L - L - 6 = 0 \), simplifying to \( L^2 + L - 6 = 0 \). Factor the quadratic equation to find \((L - 2)(L + 3) = 0\).

Step 6: Solve for L

From \((L - 2)(L + 3) = 0\), we get two possible solutions for \( L \): \( L = 2 \) or \( L = -3 \). However, we need to check which of these is consistent with the sequence's convergence.

Step 7: Determine the Valid Solution

Considering the recursive update formula \( a_{n+1} = \frac{a_n + 6}{a_n + 2} \) and \( a_1 = -1 \), we examine the behavior of the sequence. For \( a_n \to 2 \), \( a_{n+1} \to 2 \) would be consistent and stable, while moving towards \( -3 \) causes instability. Thus, the sequence converges to \( L = 2 \).

Key Concepts

Limit of a Sequence

When dealing with sequences in mathematics, one fundamental concept is the **limit of a sequence**. This describes the value that the terms of a sequence approach as the number of terms goes to infinity. It's like saying if you keep following a certain path, where do you end up?

In mathematical notation, we express a limit as \( \lim_{{n \to \infty}} a_n = L \), where \( a_n \) is the nth term of the sequence and \( L \) is the limit. For the sequence given in our exercise, we're told it converges, meaning that it approaches a specific numerical value. This entails the terms get arbitrarily close to \( L \) after some point.

When we state that this sequence converges, it implies that whatever fluctuations the terms have, they eventually settle at a particular value, which in our case is determined to be \( L = 2 \). This settlement point is what we call the sequence's limit.

Recursive Formula

In mathematics, a **recursive formula** helps define the terms of a sequence with respect to the previous terms. It serves as a rule that translates the sequence's behavior step by step.

The recursive formula for our sequence is given as \( a_{n+1} = \frac{a_n + 6}{a_n + 2} \). This means that each term in the sequence is calculated based on the previous term. Starting with \( a_1 = -1 \), we use the formula repeatedly to find the subsequent values like \( a_2, a_3, \) and so on.

• The beauty of a recursive formula is that it creates a dynamic system of progression, where each state is reliant on the previous one.
• It is a mathematical expression that outlines how to move forward from one term to the next, providing a way to systematically analyze sequences without a direct formula for the nth term.
• This type of formula is crucial for understanding many sequences as it offers insights into the underlying patterns and potential limits lying within.

By assuming that our sequence ultimately converges, we substitute this eventual limit into the recursive formula to derive a clean equation, reflecting that any fluctuations earlier in the sequence balance out at infinity.

Quadratic Equations

**Quadratic equations** form a significant part of solving for the limits in sequences like ours. A quadratic equation is of the general form \( ax^2 + bx + c = 0 \).

In our exercise, we derived the equation \( L^2 + L - 6 = 0 \) when setting the limit \( L \) into the recursive formula. The expression \( L^2 + L - 6 = 0 \) is a standard quadratic equation.

Solving quadratic equations typically involves factoring, using the quadratic formula, or completing the square. For this quadratic equation:\

• We factored it as \((L - 2)(L + 3) = 0\).
• This gives two solutions for \( L \): \( L = 2 \) and \( L = -3 \).
• Analyzing the sequence's behavior, we determine \( L = 2 \) as the valid solution since it aligns logically with the context of our sequence's convergence.

Quadratics like this are powerful tools in finding limits because they often naturally result from the algebraic manipulations of recursive sequences, allowing us to break them into manageable parts.