Textbook Question
86. This exercise explores the difference between
lim(x→∞)(1 + 1/x²)^x
and
lim(x→∞)(1 + 1/x)^x = e
c. Confirm your estimate of lim(x→∞)f(x) by calculating it with l’Hôpital’s Rule.
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.8.6c
6. Which of the following functions grow faster than ln(x) as x→∞? Which grow at the same rate as ln(x)? Which grow slower?
c. 1/√x
Verified step by step guidance1
Recall that the natural logarithm function \(\ln(x)\) grows without bound as \(x \to \infty\), but it does so very slowly compared to many other functions.
Consider the function given: \(f(x) = \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\). As \(x \to \infty\), this function approaches zero because the denominator grows without bound.
To compare growth rates, analyze the limit of the ratio \(\frac{f(x)}{\ln(x)} = \frac{1/\sqrt{x}}{\ln(x)} = \frac{1}{\sqrt{x} \cdot \ln(x)}\) as \(x \to \infty\).
Since \(\sqrt{x}\) and \(\ln(x)\) both grow to infinity, their product \(\sqrt{x} \cdot \ln(x)\) also grows to infinity, making the ratio approach zero.
Because the ratio \(\frac{f(x)}{\ln(x)} \to 0\), the function \(\frac{1}{\sqrt{x}}\) grows slower than \(\ln(x)\) as \(x \to \infty\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Growth Rates of Functions as x Approaches Infinity
Understanding how functions behave as x approaches infinity is essential for comparing their growth rates. Functions can grow faster, slower, or at the same rate depending on how quickly their values increase. For example, polynomial functions grow faster than logarithmic functions, while functions approaching zero grow slower.
Recommended video:
05:36
Integrals of Natural Exponential Functions (e^x) Example 3
Behavior of the Natural Logarithm Function ln(x)
The natural logarithm function ln(x) increases without bound as x approaches infinity, but it does so very slowly compared to polynomial or exponential functions. Recognizing the slow growth of ln(x) helps in comparing it with other functions to determine relative growth rates.
Recommended video:
05:18Derivative of the Natural Logarithmic Function
Limits and Comparison of Functions Using Limits
To compare growth rates, we often use limits of ratios of functions as x approaches infinity. If the limit of f(x)/g(x) is zero, f grows slower than g; if it is infinity, f grows faster; if it is a finite nonzero constant, they grow at the same rate. This method provides a precise way to classify growth behavior.
Recommended video:
07:45Limit Comparison Test
Related Practice
Textbook Question
Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
7. c. arcsec(-2)
Textbook Question
In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.
1. 2y' + 3y = e^(-x)
c. y = e^(-x) + Ce^(-(3/2)x)
Textbook Question
5. Which of the following functions grow faster than ln(x) as x→∞? Which grow at the same rate as ln(x)? Which grow slower?
c. ln(√x)
Textbook Question
In Exercises 1–4, solve for t.
1. c. e^((ln 0.2)t) = 0.4
Textbook Question
80. Find all values of c that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.
c. f(x) = x³/ (3 - 4x), g(x) = x², (a, b) = (0, 3)