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Chapter 14: Problem 10

Review the argument that \(\lim _{t \rightarrow \infty} \frac{e^{t}}{t}=\infty,\) Equation14.20. Use similar steps to show that $$ \lim _{t \rightarrow \infty} \frac{\ln t}{t}=0 $$ You may find the following algebra helpful. $$ \frac{\ln t}{t}=\frac{2 \ln t-\ln t}{t}=\frac{\ln t^{2}-\ln t}{t}<\frac{\ln t^{2}-\ln t}{t^{2}-t} $$ for $$ 2

Short Answer

Expert verified
\(\lim_{t \rightarrow \infty} \frac{\ln t}{t} = 0\).

Step by step solution

01

Introduction

We need to evaluate the limit \(\lim_{t \rightarrow \infty} \frac{\ln t}{t} = 0\). To do this, we'll analyze the expression \(\frac{\ln t}{t}\) and use some transformations and inequalities.
02

Transform the Expression

Given the hint in the problem, rewrite the expression as \(\frac{\ln t}{t} = \frac{2\ln t - \ln t}{t} = \frac{\ln t^2 - \ln t}{t}\).
03

Inequality Transformation

Notice that \(\frac{\ln t}{t} = \frac{\ln t^2 - \ln t}{t} < \frac{\ln t^2 - \ln t}{t^2 - t}\) for \(2<t\). We'll use this inequality to show that the original limit goes to zero.
04

Show the Limit of the Inequality

We need to determine the behavior of \(\frac{\ln t^2 - \ln t}{t^2 - t} = \frac{2\ln t - \ln t}{t^2 - t} = \frac{\ln t}{t(t-1)}\). As \(t \rightarrow \infty\), the term \(t(t-1)\) grows much faster than \(\ln t\), so the entire expression goes to zero.
05

Final Conclusion

Since \(\frac{\ln t}{t} < \frac{\ln t}{t(t-1)}\) and \(\lim_ {t \rightarrow \infty} \frac{\ln t}{t(t-1)} = 0\), it can be concluded that \(\lim_{t \rightarrow \infty} \frac{\ln t}{t} = 0\).

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Key Concepts

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

Logarithmic Functions
Logarithmic functions are an essential part of calculus. They are inverse functions of exponential functions and often appear in limits and derivatives. The natural logarithm, represented as \( \ln t \), is particularly common.
  • The natural logarithm \( \ln t \) uses base \( e \), where \( e \) is approximately 2.718.
  • Logarithms transform multiplication into addition. For example, \( \ln(a \cdot b) = \ln a + \ln b \).
  • This property is useful when manipulating expressions to simplify limits or integration.
In the context of limits, the growth rate of a logarithmic function is slower than that of a polynomial or exponential function. This is critical when determining the limit of \( \frac{\ln t}{t} \) as \( t \to \infty \). The slow growth of logarithmic functions is the key reason such limits often resolve towards zero.
Limit at Infinity
When evaluating a limit as a variable approaches infinity, we're observing the behavior of the function as the variable grows indefinitely large. In the exercise, we examined the limit \( \lim_{t \to \infty} \frac{\ln t}{t} = 0 \).
  • As \( t \to \infty \), logarithmic functions grow slower than linear terms, causing the fraction \( \frac{\ln t}{t} \) to tend toward zero.
  • The limit at infinity is calculated by understanding which part of the expression grows faster. In this example, \( t \) grows much faster than \( \ln t \).
This understanding helps us manipulate expressions to conclude that something becomes negligible as a variable becomes very large. For such problems, rewriting expressions and assessing growth rates can clarify behavior and convergence at infinity.
Asymptotic Behavior
Asymptotic behavior gives insight into how a function behaves as its input gets very large. Understanding this can be crucial, especially in calculus, when solving limits and attempting to predict function behavior.
  • In asymptotic analysis, we often compare functions to find which is growing faster.
  • The notation \( f(t) \sim g(t) \) as \( t \to \infty \) indicates \( f(t) \) and \( g(t) \) grow at similar rates.
  • This notion helps to analyze expressions like \( \frac{\ln t}{t} \), where \( t \) increases faster than \( \ln t \).
For the exercise, the asymptotic behavior highlights that despite \( \ln t \) remaining non-zero, the denominator \( t \) becomes so large that it dwarfs \( \ln t \), managing the limit toward zero. This concept is a vital tool in evaluating how functions operate under extreme values and ensuring the accuracy of limits as variables grow infinite.

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Most popular questions from this chapter

Find the first five terms of the solutions to \(\begin{array}{llll}\text { a. } & Q_{0}=1 & Q_{1}=0 & Q_{t+1}-Q_{t}=Q_{t-1} \\\ \text { b. } & Q_{0}=1 & Q_{1}=0 & Q_{t}-Q_{t-1}=Q_{t-2} \\ \text { c. } & Q_{0}=1 & Q_{1}=1 & Q_{t+1}-Q_{t}=Q_{t-1} \\ \text { d. } & Q_{0}=1 & Q_{1}=0 & Q_{t+1}=-Q_{t-1} \\ \text { e. } & Q_{0}=1 & Q_{1}=1 & Q_{t}=-Q_{t-2} \\ \text { f. } & Q_{0}=1 & Q_{1}=0 & Q_{2}=2 & Q_{t+1}-Q_{t}=Q_{t}-Q_{t-2} \\ \text { g. } & Q_{0}=1 & Q_{1}=1 & Q_{2}=1 & Q_{t+1}=Q_{t}-Q_{t-1}+Q_{t-2} \\ \text { h. } & Q_{0}=0.6 & Q_{1}=0.7 & Q_{2}=0.5 \quad Q_{t+1}-Q_{t}=0.6 Q_{t-2}\left(1-Q_{t-2}\right) \\ & & & & -0.2 Q_{t-1}\end{array}\)

Suppose there is a lake of volume 100,000 cubic meters and a stream that runs into the lake and out of the lake at a rate of 2,000 meters cube per day. Suppose further a mining operation is developed in the drainage area to the lake and one kilogram of mercury leaches into the lake each day and is mixed uniformly into the water. How much mercury is in the lake after 1000 days?

Contrast the growths per time period described by $$ P_{t+1}-P_{t}=0.2 \times P_{t} \times\left(1-\frac{P_{t}}{1000}\right) $$ (a) when \(P_{t}=900\) and \((\mathrm{b})\) when \(P_{t}=1000\) and \((\mathrm{c})\) when \(P_{t}=1100\).

Exercise 14.4 .2 Find the locally stable equilibrium points of the following iteration functions. Draw the graphs of the iteration function \(y=f(x)\) and the diagonal \(y=x .\) Start with \(x_{0}=1.0\) and show the paths of the iterates on your graphs. With \(x_{0}=1.0\) compute \(x_{1}, \cdots, x_{10}\) a. \(\quad x_{t+1}=\frac{x_{t}+2 / x_{t}}{2}\) \(\text { b. } x_{t+1}=\frac{x_{t}+5 / x_{t}}{2}\)

Show that if \(B\) is a number greater than \(1,\) and \(n\) is a positive integer, $$ \lim _{t \rightarrow \infty} \frac{B^{t}}{t^{n}}=\infty $$
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