Problem 75
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 \(n>N\). $$ \lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0 $$
Step-by-Step Solution
Verified Answer
For every \( \epsilon > 0 \), choose \( N = \left\lceil \frac{1}{\sqrt{\epsilon}} \right\rceil \) to satisfy the limit definition.
1Step 1: Understanding the Limit Expression
We need to prove that \( \lim_{n \to \infty} \frac{1}{n^2} = 0 \) using the formal definition of a limit. The formal definition states that for every \( \epsilon > 0 \), there exists a number \( N \) such that if \( n > N \), then \( \left| \frac{1}{n^2} - 0 \right| < \epsilon \).
2Step 2: Simplify the Expression
First, notice that \( \left| \frac{1}{n^2} - 0 \right| = \left| \frac{1}{n^2} \right| = \frac{1}{n^2} \). This simplifies our task to finding \( N \) such that \( \frac{1}{n^2} < \epsilon \) for all \( n > N \).
3Step 3: Find the Inequality Condition
We want \( \frac{1}{n^2} < \epsilon \). To solve for \( n \), take the reciprocal of both sides (remembering to reverse the inequality): \( n^2 > \frac{1}{\epsilon} \).
4Step 4: Determine the Value of N
Taking the square root of both sides gives \( n > \frac{1}{\sqrt{\epsilon}} \). Thus, we can choose \( N = \left\lceil \frac{1}{\sqrt{\epsilon}} \right\rceil \), where \( \lceil x \rceil \) represents the smallest integer greater than or equal to \( x \). This ensures that \( n > N \) will satisfy our condition.
Key Concepts
Epsilon-Delta DefinitionConvergence in SequencesLimits at Infinity
Epsilon-Delta Definition
The epsilon-delta (\(\epsilon\)-\(\delta\)) definition is an essential concept in calculus, particularly when discussing limits. This formal definition provides a way to rigorously prove that a function approaches a specific value as its input approaches some value. For sequences, the definition is slightly adapted. Here’s how it works:
- Given a sequence \(a_n\) that converges to \(a\), the goal is to show that for every arbitrarily small positive number \(\epsilon\), we can find an integer \(N\) such that all terms of the sequence beyond the \(N\)-th term are within \(\epsilon\) of \(a\).
- In simpler terms, we want \(\left|a_n - a\right| < \epsilon\) whenever \(n > N\).
Convergence in Sequences
Convergence in sequences describes how a sequence of numbers approaches a particular value as the number of terms increases. To say that a sequence converges, every term beyond a certain point should approximate the limit more closely.
- A sequence \(a_n\) converges to a limit \(a\) if the terms get arbitrarily close to \(a\) as \(n\) becomes very large.
- The epsilon-delta definition provides the framework to prove this convergence rigorously.
Limits at Infinity
When discussing limits at infinity, we look at the behavior of a sequence or function as its input grows infinitely large. This is crucial in understanding how sequences settle into their limiting value over time.
- The goal is to determine the value that a sequence approaches as \(n\) becomes very large.
- For the sequence \(\frac{1}{n^2}\), the exercise illustrated that as \(n\) increases infinitely, \(\frac{1}{n^2}\) approaches 0.
Other exercises in this chapter
Problem 73
In Problems \(73-78\), use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\ep
View solution Problem 74
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 77
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