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Chapter 27: Problem 7

In Exercises \(3-36,\) evaluate each limit (if it exists). Use L'Hospital's rule (if appropriate). $$\lim _{x \rightarrow \infty} \frac{x \ln x}{x+\ln x}$$

Short Answer

Expert verified
The limit is \( \infty \).

Step by step solution

01

Identify the Indeterminate Form

As \( x \to \infty \), both \( x \ln x \) and \( x + \ln x \) tend to infinity. So the expression \( \frac{x \ln x}{x + \ln x} \) takes the indeterminate form \( \frac{\infty}{\infty} \), which allows us to use L'Hospital's Rule.
02

Apply L'Hospital's Rule

L'Hospital's Rule states that for functions \( f(x) \) and \( g(x) \) where the limit \( \frac{f(x)}{g(x)} \) is of the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we have: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]Applying this, we find:- \( f(x) = x \ln x \Rightarrow f'(x) = \ln x + 1 \)- \( g(x) = x + \ln x \Rightarrow g'(x) = 1 + \frac{1}{x} \) Thus, the limit becomes:\[ \lim_{x \to \infty} \frac{\ln x + 1}{1 + \frac{1}{x}} \]
03

Simplify the Expression

As \( x \to \infty \), \( \frac{1}{x} \to 0 \). Therefore, the expression \( 1 + \frac{1}{x} \) simplifies to 1. The limit now becomes:\[ \lim_{x \to \infty} \frac{\ln x + 1}{1} = \lim_{x \to \infty} (\ln x + 1) \]
04

Final Evaluation of the Limit

As \( x \to \infty \), the term \( \ln x \to \infty \). Therefore:\[ \lim_{x \to \infty} (\ln x + 1) = \infty \]Thus, the entire limit evaluates to \( \infty \).

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Key Concepts

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

Indeterminate Forms
The concept of indeterminate forms is a crucial part of calculus, especially when dealing with limits. An expression is said to be in an indeterminate form if it lacks a defined output in its current state. This usually happens when direct substitution into a limit gives results that do not readily indicate what the limit is.
  • Common indeterminate forms include types like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), etc.
  • In this exercise, the original expression \( \frac{x \ln x}{x + \ln x} \) tends to the form \( \frac{\infty}{\infty} \) as \( x \to \infty \).
Recognizing an indeterminate form is essential because it allows the application of specific calculus techniques, such as L'Hospital's Rule, to resolve the ambiguity and find the actual limit. Understanding these forms requires us to look beyond direct computation and apply more nuanced mathematical tools.
Calculus Limits
Limits are a foundational concept in calculus that help us understand the behavior of functions as they approach a specific point. They don't focus on the value of the function at a point, but rather on what the function is approaching as the input gets closer to that point.
  • Limits are essential in defining derivatives and integrals, and they play a pivotal role in continuous mathematics.
  • In this exercise, we deal with the limit \( \lim_{x \to \infty} \), which examines the behavior of the expression as \( x \) grows without bounds.
Using limits allows for the evaluation of expressions that aren't straightforward, such as those involving \( \infty \) or 0 directly. In calculus, when faced with indeterminate forms in limits, tools like L'Hospital's Rule become invaluable for simplifying the process and achieving accurate results.
Derivatives
Derivatives are a fundamental idea in calculus, representing the rate of change of a function. When functions are expressed by their limits, we often use their derivatives to simplify and solve.
  • Derivatives provide the slope of a function at any given point, representing how the function's values change as the input changes.
  • In L'Hospital's Rule, derivatives are used to transform complicated limits into simpler expressions that are often easier to evaluate.
In this particular exercise, we derived:
  • \( f(x) = x \ln x \), leading to \( f'(x) = \ln x + 1 \)
  • \( g(x) = x + \ln x \), leading to \( g'(x) = 1 + \frac{1}{x} \)
Applying derivatives through L'Hospital's Rule here transitioned us from a complex indeterminate form to a more straightforward limit. This underscores how derivatives aren't just for determining slopes but also serve as powerful tools for evaluating limits and solving different calculus problems.

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Most popular questions from this chapter

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.37 shows the graph of \(y=\sinh x\) The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{\mu}+\mathrm{e}^{-u}\right)\) Figure 27.38 shows the graph of \(y=\cosh x\) These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\) Show that \(\frac{d}{d x} \sinh u=\cosh u \frac{d u}{d x}\) and \(\frac{d}{d x} \cosh u=\sinh u \frac{d u}{d x}\) where \(u\) is a function of \(x\)

Solve the given problems. A camera is on the starting line of a drag race \(15.0 \mathrm{m}\) from a racing car. After 1.5 s the car has traveled \(30.0 \mathrm{m}\) and the camera is rotating at 0.75 rad/s while filming the car. What is the speed of the car at this time?

Find the derivatives of the given functions. $$y=\sin (3 x+2)$$

Find the derivatives of the given functions. $$p=\frac{1}{\sin s}+\frac{1}{\cos s}$$

Solve the given problems.Find the slope of a line tangent to the curve of \(y=2 \cot 3 x\) where \(x=\pi / 12 .\) Verify the result by using the numerical derivative feature of a calculator.
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