Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problema 7.8.22

Capitolo 7, Problema 7.8.22

22. The function ln x grows slower than any polynomial Show that ln(x) grows slower as x→∞ than any nonconstant polynomial.

Guida verificata passo dopo passo

Recall the definition of growth rates: to show that \(\ln(x)\) grows slower than any nonconstant polynomial \(x^n\) (where \(n > 0\)) as \(x \to \infty\), we need to show that the ratio \(\frac{\ln(x)}{x^n}\) approaches 0 as \(x\) becomes very large.

Set up the limit to compare the growth rates: consider \(\lim_{x \to \infty} \frac{\ln(x)}{x^n}\). If this limit equals 0, it means \(\ln(x)\) grows slower than \(x^n\).

Apply L'Hôpital's Rule because the limit is of the form \(\frac{\infty}{\infty}\). Differentiate numerator and denominator with respect to \(x\): the derivative of \(\ln(x)\) is \(\frac{1}{x}\), and the derivative of \(x^n\) is \(n x^{n-1}\).

Rewrite the limit after differentiation: \(\lim_{x \to \infty} \frac{\frac{1}{x}}{n x^{n-1}} = \lim_{x \to \infty} \frac{1}{n x^n}\). Since \(n > 0\), as \(x \to \infty\), \(x^n \to \infty\), so the whole expression approaches 0.

Conclude that since \(\lim_{x \to \infty} \frac{\ln(x)}{x^n} = 0\), the logarithmic function \(\ln(x)\) grows slower than any nonconstant polynomial \(x^n\) as \(x\) approaches infinity.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Growth Rates and Asymptotic Behavior

Understanding how functions behave as their input approaches infinity is crucial. Growth rates compare how quickly functions increase, with slower-growing functions eventually becoming negligible compared to faster-growing ones. This concept helps in analyzing limits involving infinity.

Video consigliato:

Percorso guidato

04:16

Intro To Related Rates

Limits at Infinity

Limits at infinity describe the behavior of functions as the variable grows without bound. Evaluating limits like lim(x→∞) (ln x) / (x^a) helps determine which function dominates, showing that logarithmic functions grow slower than any positive power of x.

Video consigliato:

03:07

Cases Where Limits Do Not Exist

Properties of Logarithmic and Polynomial Functions

Logarithmic functions increase without bound but very slowly, while polynomial functions grow much faster as x increases. Recognizing these properties allows us to compare their growth and prove that ln(x) is dominated by any nonconstant polynomial as x→∞.

Video consigliato:

Percorso guidato

06:21

Properties of Functions

Pratica correlata

Domanda del libro di testo

Solve the differential equation in Exercises 9–22.

10. (dy/dx) = x²√y, y > 0

views

Domanda del libro di testo

Evaluate the integrals in Exercises 53–76.

75. ∫y dy/√(1-y^4)

views

Domanda del libro di testo

Evaluate the integrals in Exercises 91–102.

96. ∫dy/((arcsin y)(1-y²))

views

Domanda del libro di testo

Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.

f(x) = x⁵

views

Domanda del libro di testo

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.

119. y = (sin x)ˣ

views

Domanda del libro di testo

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x

views