Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Transform the Expression

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

Step 3: 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

Step 4: 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.

Step 5: 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 \(\lim_{t \rightarrow \infty} \frac{\ln t}{t} = 0\).

Key Concepts

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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Other exercises in this chapter

Problem 8

Show that $$ \lim _{n \rightarrow \infty} 2^{n}=\infty $$ using the following procedure. Argue using mathematical induction that all of the statements in the se

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Problem 9

Recall that the graph of \(y=\ln x\) is the reflection of the graph of \(y=e^{x}\) about the line \(y=x .\) What is \(\lim _{x \rightarrow \infty} \ln x ?\)

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Problem 10

Compute solutions until they become negative or imaginary for the systems: a. $$ P_{0}=\pi / 4 \quad \text { b. } \quad P_{0}=5 $$ $$ P_{t+1}=\ln \left(\tan \le

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Problem 11

Evaluate a. \(\lim _{t \rightarrow \infty} \frac{\ln \sqrt{t}}{\sqrt{t}}\) b. \(\lim _{t \rightarrow \infty} \frac{\ln t}{\sqrt{t}}\)

View solution

Key Concepts

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.