/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Evaluate the limit using an appr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 61

Evaluate the limit using an appropriate substitution. (See Exercises \(45-46\) of Section \(1.3 .\) ) $$\lim _{x \rightarrow+\infty} \frac{\ln 2 x}{\ln 3 x}[\text {Hint: } t=\ln x]$$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Substitute Variable

Let the hint guide us and make the substitution: let \( t = \ln x \). This implies that as \( x \to +\infty \), \( t \to +\infty \) as well. Hence, the given limit can be rewritten in terms of \( t \).
02

Substitute in the Expression

Rewrite \( \ln(2x) \) and \( \ln(3x) \) using the properties of logarithms: \( \ln(2x) = \ln 2 + \ln x = \ln 2 + t \) and \( \ln(3x) = \ln 3 + \ln x = \ln 3 + t \). Substitute these into the original limit.
03

Simplify the Limit Expression

The limit now becomes: \[ \lim_{t \to +\infty} \frac{\ln 2 + t}{\ln 3 + t}. \] Simplify this expression by dividing the numerator and denominator by \( t \): \[ = \lim_{t \to +\infty} \frac{\frac{\ln 2}{t} + 1}{\frac{\ln 3}{t} + 1}. \]
04

Evaluate the Limit

As \( t \to +\infty \), the terms \( \frac{\ln 2}{t} \) and \( \frac{\ln 3}{t} \) approach 0. Therefore, the expression simplifies to \[ \lim_{t \to +\infty} \frac{0 + 1}{0 + 1} = \frac{1}{1} = 1. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Logarithmic Functions
Logarithmic functions are an essential concept in calculus, often serving as a powerful tool for analyzing limits, series, and integrals. The logarithm has the remarkable property of turning multiplication inside its argument into addition, which makes it particularly useful when simplifying expressions.
  • For any positive number, the logarithmic function is defined as the inverse of the exponential function. This means for a given base, say base 10, the equation can be defined as: If 10 raised to the power of y is equal to x, then the logarithm of x, written as \(\log_{10}(x)\), is equal to y.
  • In the problem, we make use of natural logarithms (ln), which use the base \(e\), a mathematical constant approximately equal to 2.71828.
  • Using properties of logarithms, such as \(\ln(ab) = \ln a + \ln b\), allows for breaking down complex expressions into manageable parts. This is exactly what we do when simplifying \(\ln(2x)\) and \(\ln(3x)\).
Being able to manipulate logarithmic expressions is crucial when tackling calculus problems involving limits, differentiation, or integration.
Substitution Method
Substitution is a handy method in calculus used to simplify complex expressions and solve limits, integrals, or equations. This technique involves replacing a term or variable in the equation with another expression, making the problem easier to manage.
  • In our problem, we substitute \(t = \ln x\). This substitution helps us reframe the original limit problem, enabling us to focus on the behavior of the expression as the new variable \(t\) tends to infinity.
  • By changing the variable, we can apply known limit operations and properties more straightforwardly. For instance, simpler terms emerge, like \(\ln 2 + t\) and \(\ln 3 + t\), replacing potentially cumbersome expressions.
  • This substitution not only simplifies our calculations but also makes it more intuitive to apply the limit as \(t\) grows infinitely large.
Substitution is a versatile method, pivotal in effectively tackling problems involving exponential growth, decay, and rates in calculus.
Limits at Infinity
Limits at infinity are a key concept in calculus, representing how a function behaves as its input grows without bound. Understanding this concept is essential for studying functions and evaluating their behavior over wider domains.
  • In this problem, the goal is to assess the limits of the expression \(\frac{\ln 2 + t}{\ln 3 + t}\) as \(t\) approaches infinity.
  • When faced with limits at infinity, a useful strategy is to evaluate the dominant terms. As both the numerator and the denominator are dominated by the term \(t\), other components become negligible in comparison.
  • Dividing both parts of the fraction by \(t\) reveals the core behavior of the function, illustrating that all constant terms approach zero: \(\frac{1}{1} = 1\) is the resulting simplified limit as \(t\) increases without bound.
Understanding limits at infinity enables students to make sense of long-term behavior of functions and gain insight into their asymptotic properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Identify the domain and range of the function, and then sketch the graph of the function without using a graphing utility. (a) \(f(x)=1+\ln (x-2)\) (b) \(g(x)=3+e^{x-2}\)

Let \(f(x)=x^{6}+3 x+5, x \geq 0 .\) Show that \(f\) is an invertible function and that \(f^{-1}\) is continuous on \([5,+\infty)\).

The equation \(Q=12 e^{-0.055 t}\) gives the mass \(Q\) in grams of radioactive potassium-42 that will remain from some initial quantity after \(t\) hours of radioactive decay. (a) How many grams were there initially? (b) How many grams remain after 4 hours? (c) How long will it take to reduce the amount of radioactive potassium- 42 to half of the initial amount?

According to Ohm's law, when a voltage of \(V\) volts is applied across a resistor with a resistance of \(R\) ohms, a current of \(I=V / R\) amperes flows through the resistor. (a) How much current flows if a voltage of 3.0 volts is applied across a resistance of 7.5 ohms? (b) If the resistance varies by 卤0.1 ohm, and the voltage remains constant at 3.0 volts, what is the resulting range of values for the current? (c) If temperature variations cause the resistance to vary by \(\pm \delta\) from its value of 7.5 ohms, and the voltage remains constant at 3.0 volts, what is the resulting range of values for the current? (d) If the current is not allowed to vary by more than \(\epsilon=\pm 0.001\) ampere at a voltage of 3.0 volts, what variation of \(\pm \delta\) from the value of 7.5 ohms is allowable? (e) Certain alloys become superconductors as their temperature approaches absolute zero \(\left(-273^{\circ} \mathrm{C}\right),\) meaning that their resistance approaches zero. If the voltage remains constant, what happens to the current in a superconductor as \(R \rightarrow 0^{+} ?\)

(a) Use the Intermediate-Value Theorem to show that the equation \(x=\cos x\) has at least one solution in the interval \([0, \pi / 2]\) (b) Show graphically that there is exactly one solution in the interval. (c) Approximate the solution to three decimal places.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.