/*! 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 19 Find the limit. Use I'Hopital's ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 19

Find the limit. Use I'Hopital's rule if it applies. $$\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{\cos x}$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Check for Indeterminate Form

First, we need to substitute the limit point into the function to check if it is in an indeterminate form. Substituting 0 into the expression, we have: \( \frac{e^{4 \cdot 0} - 1}{\cos(0)} = \frac{1 - 1}{1} = \frac{0}{1} = 0 \). Since this is not an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), I'Hopital's rule does not apply.
02

Evaluate the Limit Directly

Since we found that substituting \(x = 0\) doesn't result in an indeterminate form, we can evaluate the limit directly. We have: \( \lim_{x \to 0} \frac{e^{4x} - 1}{\cos x} \). Rationalizing, the numerator simplifies to 0 (since \(e^0 = 1\)), and the denominator \( \cos x \approx 1 \) as \( x \rightarrow 0 \). Thus, the limit evaluates to \(0\).
03

Conclusion

Having determined that the form is not indeterminate, and knowing that the expression evaluates to 0 directly at \(x=0\), we conclude that the limit exists and is equal to 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.

L'Hopital's Rule
L'Hopital's Rule is a widely used tool in calculus for evaluating limits, especially when you encounter an indeterminate form. The rule is particularly handy when a straightforward evaluation doesn't give a clear result.Here's how it works:
  • If the limit results in either \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form, then L'Hopital's Rule states that the limit of a quotient can be found by taking the derivative of the numerator and the denominator separately.
  • The expression becomes: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) (if the right-hand side exists).
L'Hopital's Rule can simplify difficult limit problems, but remember it's only applicable in specific indeterminate forms.
Indeterminate Forms
In calculus, indeterminate forms are expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) that do not have a straightforward limit. These forms often appear during limit calculations.When you find yourself with any indeterminate expression, it indicates that the limit process needs additional methods to resolve. L'Hopital's Rule is often the go-to solution for dealing with these forms. However, other techniques such as algebraic manipulation or series expansion might also be needed to handle complex cases.It's crucial to first identify whether the expression is genuinely indeterminate before trying to apply any rules or techniques.
Direct Evaluation of Limits
Direct evaluation of limits is often the simplest method. Imagine you have a limit expression, like \( \lim_{x \to a} f(x) \). If substituting the value \(a\) into the function doesn't lead to any indeterminate forms, then the limit can often be evaluated directly.For example, if substituting does not present the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then it is generally safe to evaluate directly. This approach can save time and effort compared to other methods. Always start with direct evaluation, as it might solve the problem right away, just as with the given exercise where the limit directly evaluated to 0 without complications.
Limit Calculation Steps
Calculating limits effectively requires a sequence of steps to correctly approach the problem:
  • **Check for Indeterminate Forms:** Before proceeding with any calculations, try substituting the limit point in the given expression.
  • **Decide on a Method:** If an indeterminate form results, consider using L'Hopital's Rule or algebraic manipulations. Otherwise, proceed with direct evaluation.
  • **Apply Calculations:** For L'Hopital's Rule, differentiate the numerator and denominator separately. For direct evaluation, plug in the value directly.
  • **Conclude:** Analyze the results of your calculation to determine if the limit exists and what its value is.
By following a structured approach, evaluating limits can become a more straightforward task.

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

A particle moves in the \(x y\) -plane with position at time \(t\) given by \(x=\sin t\) and \(y=\cos (2 t)\) for \(0 \leq t<2 \pi.\) (a) At what time does the particle first touch the \(x\) axis? What is the speed of the particle at that time? (b) Is the particle ever at rest? (c) Discuss the concavity of the graph.

Coroners estimate time of death using the rule of thumb that a body cools about \(2^{\circ} \mathrm{F}\) during the first hour after death and about \(1^{\circ} \mathrm{F}\) for each additional hour. Assuming an air temperature of \(68^{\circ} \mathrm{F}\) and a living body temperature of \(98.6^{\circ} \mathrm{F}\), the temperature \(T(t)\) in \(^{\circ} \mathrm{F}\) of a body at a time \(t\) hours since death is given by $$T(t)=68+30.6 e^{-k t}$$ (a) For what value of \(k\) will the body cool by \(2^{\circ} \mathrm{F}\) in the first hour? (b) Using the value of \(k\) found in part (a), after how many hours will the temperature of the body be decreasing at a rate of \(1^{\circ} \mathrm{F}\) per hour? (c) Using the value of \(k\) found in part (a), show that, 24 hours after death, the coroner's rule of thumb gives approximately the same temperature as the formula.

Determine whether the limit exists, and where possible evaluate it. $$\lim _{x \rightarrow 0^{+}} x \ln x$$

Find the speed for the given motion of a particle. Find any times when the particle comes to a stop. $$x=t^{2}-4 t, y=t^{3}-12 t$$

Explain why 1 'Hopital's rule cannot be used to calculate the limit. Then evaluate the limit if it exists. $$\lim _{x \rightarrow \infty} \frac{e^{-x}}{\sin x}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

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.