Problem 61 - Calculus for Biology and Medicine

Question

In Problems 61-72, \(\lim _{n \rightarrow \infty} a_{n}=a\). Find the limit \(a\), and determine \(N\) so that \(\left|a_{n}-a\right|<\epsilon\) for all \(n>N\) for the given value of \(\epsilon\) $$ a_{n}=\frac{1}{n}, \epsilon=0.01 $$

Step-by-Step Solution

Answer

The limit is 0 and take \(N = 100\). For \(n > 100\), \(|a_n - 0| < 0.01\).

Step 1: Find the Limit

The sequence is given by \( a_n = \frac{1}{n} \). To find \( \lim_{n \to \infty} a_n \), note that as \( n \) becomes very large, the value of \( \frac{1}{n} \) approaches 0. Hence, \( a = 0 \).

Step 2: Write the Inequality

We need \( |a_n - a| < \epsilon \). Substitute \( a = 0 \), \( a_n = \frac{1}{n} \), and \( \epsilon = 0.01 \). The inequality becomes \( \left| \frac{1}{n} - 0 \right| < 0.01 \).

Step 3: Solve the Inequality

The inequality \( \frac{1}{n} < 0.01 \) needs to be solved for \( n \). Solving \( \frac{1}{n} < 0.01 \) gives \( n > \frac{1}{0.01} \). This simplifies to \( n > 100 \).

Step 4: Determine Minimum N

From the inequality, we have determined \( n > 100 \). Therefore, \( N \) can be taken to be 100. For all \( n > N \), \( \left| a_n - a \right| < \epsilon \).

Key Concepts

Understanding Convergence

Convergence of a sequence refers to the idea that as we increase the value of the term index (usually represented by "n"), the values of the sequence terms get closer and closer to a specific number—known as the "limit". This limit is often denoted by "a". For instance, in our exercise, we have the sequence given by \[a_n = \frac{1}{n}\]As "n" becomes very large, the value of \(\frac{1}{n}\) gets closer to 0. In mathematical terms, we say this sequence converges to the limit 0 as \( n \to \infty \).

• Convergence means the values are getting closer to the limit.
• The limit "a" is the value the sequence approaches.
• In our sequence, \( \lim_{n \to \infty} a_n = 0 \).

By understanding convergence, it becomes easier to predict behavior of sequences as indices (like \( n \)) grow indefinitely large.

Epsilon-Delta Definition of a Limit

The epsilon-delta definition provides a formal way to define what it means when we say a sequence approaches a limit. "Epsilon" (\(\epsilon\)) represents any small positive number. We want to find a corresponding index such that if our index is beyond this number, the difference between the sequence term and the limit is less than \(\epsilon\). In simpler terms, this means the sequence terms are really close to the limit by an amount smaller than \(\epsilon\).In our example, we start with the inequality:\[ |a_n - a| < \epsilon \]Where: - \( a = 0 \) and \( a_n = \frac{1}{n} \), so \( |\frac{1}{n} - 0| < \epsilon \).With \( \epsilon = 0.01 \) in our problem, it implies finding a point \( N \) such that for any \( n > N \), this requirement of closeness is satisfied. This approach guarantees the sequence reaches and stays close to its limit beyond a certain index. This definition is pivotal for rigorously proving limits in mathematics.

Mastering Inequality Solving in Limits

Inequality solving is critical when working with sequences and their limits. Here, we use it to determine the necessary "n" for which the sequence term gets sufficiently close to the limit, as dictated by \(\epsilon\). Solving inequalities helps us find this boundary value \(N\). Consider the inequality from our example:\[ \frac{1}{n} < 0.01 \] To solve, we manipulate it algebraically:

• Multiply both sides by \(n\) to clear the fraction: \(1 < 0.01 \cdot n\).
• Divide by 0.01 to isolate \(n\): \(n > \frac{1}{0.01}\).
• Calculate \(\frac{1}{0.01}\): which equals 100.

Thus, \(n > 100\). So, whenever \(n\) is greater than 100, our sequence's terms are within the bounds set by \(\epsilon\). Therefore, for our problem, \(N = 100\) ensures the inequality holds, highlighting the importance of algebraic skills in deducing convergence scenarios.