Problem 77
Question
Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists an \(N\) such that \(\left|a_{n}-a\right|<\epsilon\) whenever \(n>N\). $$ \lim _{n \rightarrow \infty} 2^{-3 n}=0 $$
Step-by-Step Solution
Verified Answer
Choose \(N = \left\lceil \frac{\ln(\epsilon)}{-3 \ln(2)} \right\rceil\) for \(n > N\) to satisfy \(2^{-3n} < \epsilon\).
1Step 1: Understand the Limit Definition
According to the definition of a limit, we need to show that for every \( \epsilon > 0 \), there exists a natural number \( N \) such that if \( n > N \), then \( |a_n - a| < \epsilon \). In this problem, \( a_n = 2^{-3n} \) and \( a = 0 \), so we want \( |2^{-3n} - 0| < \epsilon \), which simplifies to \( 2^{-3n} < \epsilon \).
2Step 2: Determine the Inequality
We start with the inequality \( 2^{-3n} < \epsilon \). By taking the natural logarithm on both sides, we get \( \ln(2^{-3n}) < \ln(\epsilon) \). Since \( \ln(2^{-3n}) = -3n \ln(2) \), this translates to \( -3n \ln(2) < \ln(\epsilon) \).
3Step 3: Solve for n
From \( -3n \ln(2) < \ln(\epsilon) \), we solve for \( n \). First, divide both sides by \( -3\ln(2) \), remembering that this reverses the inequality direction: \[ n > \frac{\ln(\epsilon)}{-3 \ln(2)}. \]
4Step 4: Find N based on Epsilon
To ensure \( n > N \) satisfies the inequality, we choose \( N = \left\lceil \frac{\ln(\epsilon)}{-3 \ln(2)} \right\rceil \). The ceiling function \( \lceil x \rceil \) yields the smallest integer greater than or equal to \( x \), ensuring \( n \) is a natural number.
5Step 5: Verify the Result
Finally, check that with this choice of \( N \), for any \( n > N \), the original inequality \( |2^{-3n} - 0| < \epsilon \) holds true, confirming the condition of the limit definition.
Key Concepts
Understanding Inequality in Limit DefinitionRole of Natural Logarithm in Solving InequalitiesCeiling Function: Ensuring Natural Number Solutions
Understanding Inequality in Limit Definition
Inequalities play a crucial role in the formal definition of limits, guiding us to bound a sequence or function within a specified range. In the context of limits, we often deal with expressions like \( |a_n - a| < \epsilon \). This notation signifies that the difference between the sequence \( a_n \) and the limit \( a \) becomes arbitrarily small as \( n \) becomes large.
For the exercise \( \lim_{n \to \infty} 2^{-3n} = 0 \), we start by writing the inequality \( |2^{-3n} - 0| < \epsilon \), which simplifies to \( 2^{-3n} < \epsilon \). This means we want \( 2^{-3n} \), which describes the sequence, to be smaller than any positive number \( \epsilon \).
Recognizing and manipulating such inequalities is the key to confirming the behavior of a sequence as \( n \to \infty \). In practice, inequalities provide a bound, ensuring the sequence elements are within an acceptable range to satisfy the limit condition.
For the exercise \( \lim_{n \to \infty} 2^{-3n} = 0 \), we start by writing the inequality \( |2^{-3n} - 0| < \epsilon \), which simplifies to \( 2^{-3n} < \epsilon \). This means we want \( 2^{-3n} \), which describes the sequence, to be smaller than any positive number \( \epsilon \).
Recognizing and manipulating such inequalities is the key to confirming the behavior of a sequence as \( n \to \infty \). In practice, inequalities provide a bound, ensuring the sequence elements are within an acceptable range to satisfy the limit condition.
Role of Natural Logarithm in Solving Inequalities
The natural logarithm, denoted as \( \ln(x) \), is a fundamental concept in calculus that helps solve exponent-related inequalities in limit problems. When we encounter exponential terms like \( 2^{-3n} < \epsilon \), taking the natural logarithm is a powerful technique to linearize and simplify these expressions.
By applying \( \ln \) on both sides of \( 2^{-3n} < \epsilon \), we transform the inequality into a more manageable form: \( \ln(2^{-3n}) < \ln(\epsilon) \). Recognizing that \( \ln(2^{-3n}) = -3n \ln(2) \), we achieve \( -3n \ln(2) < \ln(\epsilon) \).
The beauty of using \( \ln(x) \) is seen in its properties, particularly the ability to convert multiplicative relationships into additive ones, which is critical for solving for \( n \). Knowing these transformations helps in breaking down complex limit problems into simpler algebraic forms that are easier to tackle.
By applying \( \ln \) on both sides of \( 2^{-3n} < \epsilon \), we transform the inequality into a more manageable form: \( \ln(2^{-3n}) < \ln(\epsilon) \). Recognizing that \( \ln(2^{-3n}) = -3n \ln(2) \), we achieve \( -3n \ln(2) < \ln(\epsilon) \).
The beauty of using \( \ln(x) \) is seen in its properties, particularly the ability to convert multiplicative relationships into additive ones, which is critical for solving for \( n \). Knowing these transformations helps in breaking down complex limit problems into simpler algebraic forms that are easier to tackle.
Ceiling Function: Ensuring Natural Number Solutions
The ceiling function, written as \( \lceil x \rceil \), is an essential mathematical tool when working with limits and inequalities involving natural numbers. This function elevates a real number \( x \) to the smallest integer greater than or equal to \( x \).
In the context of finding \( N \) such that \( n > N \) for a limit definition, the ceiling function ensures \( N \) is a natural number. For instance, after calculating \( n > \frac{\ln(\epsilon)}{-3 \ln(2)} \), we set \( N = \lceil \frac{\ln(\epsilon)}{-3 \ln(2)} \rceil \). This ensures that \( N \) is an integer, meeting the requirement that \( n \) be naturally greater than any fractional value of \( \frac{\ln(\epsilon)}{-3 \ln(2)} \).
Using the ceiling function is vital because it guarantees that our solution not only satisfies the mathematical constraints but also aligns with the nature of sequences and limits, which often relate to discrete values like natural numbers.
In the context of finding \( N \) such that \( n > N \) for a limit definition, the ceiling function ensures \( N \) is a natural number. For instance, after calculating \( n > \frac{\ln(\epsilon)}{-3 \ln(2)} \), we set \( N = \lceil \frac{\ln(\epsilon)}{-3 \ln(2)} \rceil \). This ensures that \( N \) is an integer, meeting the requirement that \( n \) be naturally greater than any fractional value of \( \frac{\ln(\epsilon)}{-3 \ln(2)} \).
Using the ceiling function is vital because it guarantees that our solution not only satisfies the mathematical constraints but also aligns with the nature of sequences and limits, which often relate to discrete values like natural numbers.
Other exercises in this chapter
Problem 75
Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists
View solution Problem 76
Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists
View solution Problem 78
Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists
View solution Problem 79
In Problems 79-90, use the limit laws to determine \(\lim _{n \rightarrow \infty} a_{n}=a .\) $$ \lim _{n \rightarrow \infty}\left(\frac{1}{n}+\frac{2}{n^{2}}\r
View solution