Chapter 10: Problem 92
In Exercises \(91-98\) , assume that each sequence converges and find its limit. $$ a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2} $$
Short Answer
Expert verified
The limit of the sequence is 2.
Step by step solution
01
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 \( n \to \infty \).
02
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.
03
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 \( L \).
04
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 \( L^2 + 2L = L + 6 \).
05
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 \((L - 2)(L + 3) = 0\).
06
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.
07
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 \( L = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 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.
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:\
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.