/*! 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 21 Find the limits in Problems 1-60... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 21

Find the limits in Problems 1-60; not all limits require use of l'H么pital's rule. $$ \lim _{x \rightarrow \infty} \frac{(\ln x)^{2}}{x^{2}} $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify the Form of the Limit

The limit \( \lim_{x \to \infty} \frac{(\ln x)^2}{x^2} \) is presented. As \(x\) approaches infinity, both the numerator \((\ln x)^2\) and denominator \(x^2\) approach infinity. This is an \(\frac{\infty}{\infty}\) indeterminate form, suggesting the potential use of l'H么pital's Rule.
02

Apply l'H么pital's Rule

Since the limit is in the \(\frac{\infty}{\infty}\) form, we can apply l'H么pital's Rule. Differentiate the numerator and the denominator separately: - Derivative of the numerator: \(\frac{d}{dx}(\ln x)^2 = 2(\ln x) \cdot \frac{1}{x} = \frac{2\ln x}{x}\)- Derivative of the denominator: \(\frac{d}{dx}(x^2) = 2x\)Applying l'H么pital's Rule, the limit becomes:\[ \lim_{x \to \infty} \frac{\frac{2\ln x}{x}}{2x} = \lim_{x \to \infty} \frac{2\ln x}{2x^2} = \lim_{x \to \infty} \frac{\ln x}{x^2} \]
03

Reapply l'H么pital's Rule if Necessary

The new limit \( \lim_{x \to \infty} \frac{\ln x}{x^2} \) is again an \(\frac{\infty}{\infty}\) form.- Derivative of the numerator: \(\frac{d}{dx} \ln x = \frac{1}{x}\)- Derivative of the denominator: \(\frac{d}{dx} (x^2) = 2x\)Reapply l'H么pital's Rule:\[ \lim_{x \to \infty} \frac{\frac{1}{x}}{2x} = \lim_{x \to \infty} \frac{1}{2x^2} \]
04

Evaluate the Final Limit

The expression \( \lim_{x \to \infty} \frac{1}{2x^2} \) needs to be evaluated. As \(x\) approaches infinity, the denominator \(2x^2\) becomes very large, causing the whole fraction to approach zero. Thus, the limit is:\[ 0 \]

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.

Understanding Limits in Calculus
In calculus, a limit describes the value that a function approaches as the input approaches some point. It's like getting closer and closer to a destination but never really arriving.
For instance, in the problem of finding \( \lim_{x \rightarrow \infty} \frac{(\ln x)^{2}}{x^{2}} \), we want to know what value the function approaches as \(x\) becomes infinitely large.
Limits help us understand the behavior of functions at points where they aren't defined or as inputs grow very large or very small.
  • They are foundational in defining derivatives and integrals.
  • They handle scenarios where straightforward computation is impossible.
As we analyze the limit given in the exercise, we see it leads to a special form known as an indeterminate form.
Identifying and Handling Indeterminate Forms
Indeterminate forms arise in calculus when we have expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where it's unclear what the limit should be. They signal the need for more refined techniques.
When dealing with \( \lim_{x \rightarrow \infty} \frac{(\ln x)^2}{x^2} \), you notice both parts grow without bound. This is a classic \( \frac{\infty}{\infty} \) form.
To resolve such cases, we can use l'H么pital's Rule, which provides a way to evaluate these limits by differentiating the numerator and the denominator.
  • First, differentiate the top and bottom separately.
  • Apply l'H么pital's Rule repeatedly if another indeterminate form emerges.
  • Eventually, the expression simplifies to a determinate form, making the limit clear.
Understanding and working with indeterminate forms is crucial for mastering higher-level calculus.
Applying Calculus Concepts and l'H么pital's Rule
In calculus, l'H么pital's Rule is a powerful tool for evaluating limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It allows us to transform complex limits into simpler ones by differentiating the numerator and the denominator.
Applying this rule to \( \lim_{x \to \infty} \frac{(\ln x)^{2}}{x^{2}} \) involves several steps:
  • First, differentiate \((\ln x)^2\) to get \( \frac{2\ln x}{x} \).
  • Then, differentiate \(x^2\) to get \(2x\).
  • Keep applying l'H么pital's Rule until the expression is no longer indeterminate.
In our exercise, after reapplying the rule, we eventually simplify to \( \frac{1}{2x^2} \), which clearly approaches zero as \(x\) goes to infinity.
By mastering concepts like l'H么pital's Rule, students can tackle complex calculus problems with confidence.

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

Consider the following discrete logistic model for the change in the size of a population over time: $$N_{t+1}=R_{0} N_{t}-\frac{1}{100} N_{t}^{2}$$ for \(t=0,1,2, \ldots\) (a) Find all equilibria when \(R_{0}=3.5\) and calculate which (if any) are stable. (b) Calculate the first ten terms of the sequence when \(N_{0}=10\) and describe what you see.

Find the general solution of the differential equation. $$ \frac{d y}{d t}=1-e^{-2 t}, t \geq 0 $$

The Ricker model was introduced by Ricker (1954) as an alternative to the discrete logistic equation to describe the density-dependent growth of a population. Under the Ricker model the population \(N_{t}\) sampled at discrete times \(t=0,1,2, \ldots\) is modeled by a recurrence equation $$N_{t+1}=R_{0} N_{t} \exp \left(-a N_{t}\right)$$ where \(R_{0}\) and \(a\) are positive constants that will vary between different species and between different habitats. (a) Explain why you would expect \(R_{0}>1\) (Hint: consider the population growth when \(N_{t}\) is very small.) (b) Show that the recursion relation has two equilibria, a trivial equilibrium (that is, \(N=0\) ) and another equilibrium, which you should find. (c) Show that if \(R_{0}>1\) then use the stability criterion for equilibria to show that the trivial equilibrium point is unstable. (d) Use the stability criterion for equilibria to show that the nontrivial equilibrium point is stable if \(0<\ln R_{0}<2\). (e) If \(R_{0}>1\) then \(\ln R_{0}>0\), so most populations will meet the first inequality condition. What happens if \(\ln R_{0}>2 ?\) Let's try some explicit values: \(R_{0}=10, a=1, N_{0}=1 .\) Calculate the first ten terms of the sequence, and describe in words how the sequence behaves.

Find the general antiderivative of the given function. $$ f(x)=5 e^{3 x}-\sec ^{2}(x-3) $$

(a) Find all equilibria of $$ N_{t+1}=0.7 N_{t}, \quad t=0,1,2, \ldots $$ (b) Use cobwebbing to determine the stability of the equilibria you found in (a).
See all solutions

Recommended explanations on Biology Textbooks

Bioenergetics

Read Explanation

Reproduction

Read Explanation

Biology Experiments

Read Explanation

Heredity

Read Explanation

Biological Structures

Read Explanation

Chemistry of Life

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.