/*! 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 7 Consider the function \(g(x)=x^{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the function \(g(x)=x^{2 x}\), which is defined for all \(x>0\). Observe that \(\lim _{x \rightarrow 0^{+}} g(x)\) is indeterminate due to its form of \(0^{0}\). (Think about how we know that \(0^{k}=0\) for all \(k>0\), while \(b^{0}=1\) for all \(b \neq 0\), but that neither rule can apply to \(0^{0}\).) a. Let \(h(x)=\ln (g(x))\). Explain why \(h(x)=2 x \ln (x)\). b. Next, explain why it is equivalent to write \(h(x)=\frac{2 \ln (x)}{\frac{1}{x}}\). c. Use L'Hôpital's Rule and your work in (b) to compute \(\lim _{x \rightarrow 0^{+}} h(x)\). d. Based on the value of \(\lim _{x \rightarrow 0^{+}} h(x)\), determine \(\lim _{x \rightarrow 0^{+}} g(x)\).

Short Answer

Expert verified
\(\lim_{x \to 0^+} g(x) = 1\)

Step by step solution

01

Express the function using natural logarithm

Let the function be defined as \(g(x) = x^{2x}\). Define the new function \(h(x)\) such that \(h(x) = \ln(g(x))\). Since \(g(x) = x^{2x}\), taking the natural logarithm of both sides gives us \(h(x) = \ln(x^{2x})\).
02

Simplify the logarithmic expression

Using the logarithm property \(\ln(a^b) = b \ln(a)\), we can rewrite \(h(x) = \ln(x^{2x})\) as \(h(x) = 2x \ln(x)\).
03

Rewrite using a fraction form

To apply L'Hôpital's Rule, we need to rewrite \(h(x) = 2x \ln(x)\) in a form that highlights the indeterminate nature of the limit. Rewrite it as \(h(x) = \frac{2 \ln(x)}{\frac{1}{x}}\). This is to show the limit's form is \(\frac{-\infty}{\infty}\), which is suitable for L'Hôpital's Rule.
04

Apply L'Hôpital's Rule

To find \(\lim_{x \to 0^+} h(x)\), apply L'Hôpital's Rule: \[ \lim_{x \to 0^+} \frac{2 \ln(x)}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{ d/dx (2 \ln(x))}{d/dx (\frac{1}{x})} \]. Calculate the derivatives of the numerator and the denominator. The derivative of \(2 \ln(x)\) is \(\frac{2}{x}\) and the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\).
05

Simplify the result after differentiation

Substitute the derivatives back into L'Hôpital's expression: \[ \lim_{x \to 0^+} \frac{\frac{2}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} \frac{2x^2}{-x} = \lim_{x \to 0^+} -2x \]. As \(x\) approaches 0, \(-2x\) also approaches 0, so \[ \lim_{x \to 0^+} h(x) = 0 \].
06

Determine the original limit

Since \(h(x) = \ln(g(x))\) and we have found \(\lim_{x \to 0^+} h(x) = 0\), we can use the exponential function to determine \(\lim_{x \to 0^+} g(x)\). Thus, \[ \lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} e^{h(x)} = e^0 = 1 \].

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 91Ó°ÊÓ!

Key Concepts

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

L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus for evaluating limits of indeterminate forms. When faced with an expression like \(\frac{0}{0}\) or \(\frac{\text{infinity}}{\text{infinity}}\), we can differentiate the numerator and the denominator separately, then take the limit of the resulting fraction. This allows us to simplify complex limits.
In the given exercise, we encounter the indeterminate form \(\frac{\text{-infinity}}{\text{infinity}}\) when evaluating \(h(x) = \frac{2 \text{ln}(x)}{\frac{1}{x}}\). By applying L'Hôpital’s Rule, we differentiate the numerator, \(2 \text{ln}(x)\), and the denominator, \(\frac{1}{x}\), to transform it into a simpler expression that we can evaluate as \(x\) approaches 0.
Logarithmic Differentiation
Logarithmic differentiation is a method used to take derivatives of functions of the form \(g(x) = x^{f(x)}\) by employing logarithms. This technique simplifies the differentiation of more complicated expressions by transforming them into a more manageable form.
In the problem, we start by expressing \(g(x) = x^{2x}\) in natural logarithmic terms. By defining \(h(x) = \text{ln}(g(x))\), we apply the property \(\text{ln}(a^b) = b \text{ln}(a)\) to rewrite \(g(x)\) as \(\text{ln}(x^{2x}) = 2x \text{ln}(x)\). This simplification is crucial for applying L'Hôpital's Rule later.
Exponential Function
The exponential function, denoted as \(e^x\), is a mathematical function that grows rapidly and appears frequently in calculus and complex numbers. It is the inverse function of the natural logarithm and has unique properties that make it useful in various fields.
In the context of the exercise, after computing \(\text{lim}_{x \rightarrow 0^{+}} h(x)\), we need to switch back to the original function \(g(x)\). Since \(h(x) = \text{ln}(g(x))\), we use the exponential function to solve for \(g(x)\). The limit, \(\text{lim}_{x \rightarrow 0^{+}} g(x) = e^{h(x)}\), evaluates to \(e^0 = 1\).
Indeterminate Forms
Indeterminate forms are expressions in calculus where it is not immediately clear what the limit is. Typical examples include \(\frac{0}{0}\), \(0^0\), \(\frac{\text{infinity}}{\text{infinity}}\), and others.
In the exercise, \(g(x) = x^{2x}\) creates an indeterminate form of \(0^0\) as \(x\) approaches 0 from the positive direction. By converting this into a limit problem with natural logarithms, we can use L'Hôpital’s Rule to resolve the indeterminate form and find the limit.
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It has two main branches, differential calculus, concerning slopes of curves (derivatives), and integral calculus, concerning the area under curves (integrals).
This exercise showcases some core concepts of calculus: using logarithmic differentiation to simplify complex functions and applying L'Hôpital's Rule to find limits of indeterminate forms. Mastering these techniques is essential for advancing in more sophisticated areas of mathematics and its applications in science and engineering.

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

Find the derivative of \(f(x)=9 x \sin (2 x)\) \(f^{\prime}(x)=\) _________

Find the derivative of the function \(h(r)\), below. It may be to your advantage to simplify first. $$ h(r)=\frac{r^{2}}{15 r+11} $$ \(h^{\prime}(r)=\) _______

Consider the functions \(r(t)=t^{t}\) and \(s(t)=\arccos (t),\) for which you are given the facts that \(r^{\prime}(t)=t^{t}(\ln (t)+1)\) and \(s^{\prime}(t)=-\frac{1}{\sqrt{1-t^{2}}} .\) Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain \(0

Find the derivative of \(y=\sqrt{e^{-5 t^{2}+5}}\) \(\frac{d y}{d t}=\) ________

Let \(f(v)\) be the gas consumption (in liters/km) of a car going at velocity \(v\) (in km/hour). In other words, \(f(v)\) tells you how many liters of gas the car uses to go one kilometer if it is traveling at \(v\) kilometers per hour. In addition, suppose that \(f(80)=0.05\) and \(f^{\prime}(80)=\) \(0.0004 .\) a. Let \(g(v)\) be the distance the same car goes on one liter of gas at velocity \(v\). What is the relationship between \(f(v)\) and \(g(v)\) ? Hence find \(g(80)\) and \(g^{\prime}(80)\). b. Let \(h(v)\) be the gas consumption in liters per hour of a car going at velocity \(v\). In other words, \(h(v)\) tells you how many liters of gas the car uses in one hour if it is going at velocity \(v\). What is the algebraic relationship between \(h(v)\) and \(f(v)\) ? Hence find \(h(80)\) and \(h^{\prime}(80)\). c. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.

See all solutions

Recommended explanations on Math Textbooks

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.