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Chapter 2: Problem 70

\(\lim _{n \rightarrow \infty} a_{n}=a\). Find the limit \(a\), and determine \(N\) so that \(\left|a_{n}-a\right|<\epsilon\) for all \(n>N\) for the given value of \(\epsilon\) $$ a_{n}=\ln \left(1+\frac{1}{n}\right), \epsilon=0.1 $$

Short Answer

Expert verified
The limit is \(a = 0\) and \(N = 10\) ensures \(|a_n - a| < 0.1\).

Step by step solution

01

Identify the Form of the Sequence

We are given the sequence \(a_{n} = \ln(1 + \frac{1}{n})\). We need to find the limit of this sequence as \(n\) approaches infinity.
02

Determine the Limit

Evaluate the limit: \(\lim_{n \to \infty} \ln(1 + \frac{1}{n})\). As \(n\) becomes very large, \(\frac{1}{n}\) approaches zero, which means the expression \(1 + \frac{1}{n}\) approaches 1. Therefore, the limit of \(\ln(1 + \frac{1}{n})\) as \(n\) approaches infinity is \(\ln(1) = 0\). Thus, \(a = 0\).
03

Understand the Condition \(|a_n - a| < \epsilon\)

We need to ensure that \(|a_n - a| < \epsilon\) for all \(n > N\). In simpler terms, we have \(|\ln(1 + \frac{1}{n}) - 0| < 0.1\). Since \(|\ln(1 + \frac{1}{n})| = \ln(1 + \frac{1}{n})\) for \(n > 0\), we need \(\ln(1 + \frac{1}{n}) < 0.1\).
04

Solve for \(N\)

We solve the inequality \(\ln(1 + \frac{1}{n}) < 0.1\) by exponentiating both sides: \(1 + \frac{1}{n} < e^{0.1}\). Calculate \(e^{0.1} \approx 1.10517\). Thus, \(1 + \frac{1}{n} < 1.10517\) implies \(\frac{1}{n} < 0.10517\), which simplifies to \(n > \frac{1}{0.10517} \approx 9.51\). We choose \(N = 10\) to meet this condition for all integer \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence of Sequences
When we talk about the convergence of sequences in calculus, we are exploring whether a sequence of numbers "settles down" to a single number as it progresses to infinity. If it does, it is convergent. Otherwise, it's divergent.

- **What is a Sequence?**A sequence is a set of numbers arranged in a specific order. Typically, this order follows a rule or a formula, like our sequence here: \(a_n = \ln(1 + \frac{1}{n})\).

- **Convergence Defined**A sequence \(a_n\) converges to a limit \(a\) if for any small positive number \(\epsilon\), there's a point in the sequence after which all following terms are within \(\epsilon\) of \(a\).

In our exercise, as \(n\) approaches infinity, \(\ln(1 + \frac{1}{n})\) approaches 0. Thus, the limit \(a\) is 0, exemplifying convergence to this point.

Understanding convergence is crucial for analyzing behaviors in calculus and determining limits of functions and sequences under different conditions.
Natural Logarithm
The natural logarithm, denoted \(\ln\), is a unique and essential logarithm in mathematics with the base \(e\), the Euler's number, approximately equal to 2.71828.

- **Understanding \(\ln(x)\)**The natural logarithm \(\ln(x)\) represents the power to which \(e\) must be raised to obtain \(x\). For example, if \(\ln(x) = y\), then \(e^y = x\).

- **Natural Logarithm in Sequences**In our specific sequence \(a_n = \ln(1 + \frac{1}{n})\), as \(n\) increases, \(\frac{1}{n}\) shrinks towards zero. The logarithmic function lets us examine subtle changes as the values creep toward unity, influencing the limit result.

- **Properties of \(\ln(x)\)**
  • \(\ln(1) = 0\)
  • \(\ln(ab) = \ln(a) + \ln(b)\)
  • \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\)
  • \(\ln(a^b) = b \cdot \ln(a)\)
A strong grasp of these properties helps to solve limits and analyze exponential growth and decay behavior in various fields.
Epsilon-Delta Definition of Limit
The epsilon-delta definition is a foundation in calculus for formally defining limits. It rigorously explains what it means for a sequence or a function to approach a certain value, combining two principal ideas: the distance (\(\epsilon\)) and a threshold (\(\delta\)).

- **Core Idea**A sequence \(a_n\) approaches a limit \(a\) when, for any degree of closeness we want (\(\epsilon > 0\)), there's a certain point in the sequence from which onward all terms of the sequence are within \(\epsilon\) of \(a\). This point is denoted \(N\).

- **Relating to the Exercise**Here, \(|a_n - a| < \epsilon\) translates to \(\ln(1 + \frac{1}{n}) < 0.1\) for all \(n > N\) in our problem. Solving for \(N\) involved strategic inequality and understanding exponential functions, leading to \(N = 10\) for \(\epsilon = 0.1\).

This definition ensures predictability and precision in studies involving limits, crucial for both theoretical and practical applications in mathematics and beyond.

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Most popular questions from this chapter

Hormone Implant You are studying an implanted contraceptive that releases hormone continuously into a patient's blood. The data in this question are from Nilsson et al. (1986). The device adds \(20 \mu \mathrm{g}\) of hormone to the blood each day. In the blood the hormone has first order elimination kinetics; \(4 \%\) of the hormone is eliminated each day. (a) Let the amount of hormone in the blood on day \(t\) be \(a_{t} .\) Write a word equation for the change in \(a_{t}\) over one day. (b) Put in mathematical formulas for each of the terms in your word equation from (a). (c) Assume that on day 0 no hormone is present in the patient's blood, in other words, \(a_{0}=0 .\) Use your equation from (b) to compute the amount of hormone in the blood on days \(1,2,3,4\), \(5,6 .\) (d) Over time the level of hormone in the blood converges to a limit. Find the value of this limit by looking for a fixed point of your recurrence relation in (b).

In Problems 111-118, assume that \(\lim _{n \rightarrow \infty} a_{n}\) exists. Find all fixed points of \(\left\\{a_{n}\right\\}\), and use a table or other reasoning to guess which fixed point is the limiting value for the given initial condition. $$ a_{n+1}=\frac{1}{3} a_{n}+\frac{4}{3}, a_{0}=0 $$

Model painkillers that are absorbed into the blood from a slow release pill. Ourmathematical model for the amount, \(a_{t}\), of drug in the blood t hours after the pill is taken must include the amount absorbed from the pill each hour. Our model starts with the word equation. $$ \begin{array}{c} a_{t+1}=a_{t}+\begin{array}{l} \text { amount absorbed } \\ \text { from the pill } \end{array}-\begin{array}{l} \text { amount eliminated } \\ \text { from the blood } \end{array} \end{array} $$ Assume the amount absorbed from the pill between time \(t\) and time \(t+1\) is \(10 \cdot(0.4)^{t}\). (a) The drug has first order elimination kinetics. \(10 \%\) of the drug is eliminated from the blood each hour. Write down the recursion relation for \(a_{t+1}\) in terms of \(a_{t}\) (b) Assuming that \(a_{0}=0\), meaning that no drug is present in the blood initially, calculate the amount of drug present at times \(t=1,2, \ldots, 6\) (c) What is the maximum amount of drug present at any time in this interval? At what time is this maximum amount reached? (d) Use a spreadsheet to calculate the amount of drug present in hourly intervals from \(t=0\) up to \(t=24\). (e) Show when \(t\) is large, the amount of drug present in the blood decreases approximately exponentially with \(t .\) Hint: Plot the values that you computed for \(a_{t}\) against \(t\) on semilogarithmic axes.

The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=-2 a_{n}, a_{0}=1 $$

Assume that the population growth is described by the Beverton-Holt model. Find all fixed points. \(N_{t+1}=\frac{3 N_{t}}{1+N_{t} / 100}\)
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