Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically.
![**27. \(\lim_{{x \to 0^+}} x^x\)**
---
**Explanation:**
To compute the limit \(\lim_{{x \to 0^+}} x^x\), we need to consider the behavior of the function \(f(x) = x^x\) as \(x\) approaches \(0\) from the positive side.
### Approach:
1. Consider using the natural logarithm to simplify the expression:
\[
y = x^x \implies \ln y = \ln (x^x) = x \ln x
\]
2. Now, evaluate the limit of \(\ln y\) as \(x\) approaches 0 from the right:
\[
\lim_{{x \to 0^+}} \ln y = \lim_{{x \to 0^+}} (x \ln x)
\]
3. Use L'Hôpital's Rule to evaluate this limit:
\[
\lim_{{x \to 0^+}} x \ln x \text{ is of the form } 0 \times -\infty, \text{ which we can rewrite as } \frac{\ln x}{\frac{1}{x}}
\]
Applying L'Hôpital's Rule:
\[
\lim_{{x \to 0^+}} \frac{\ln x}{1/x} = \lim_{{x \to 0^+}} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(1/x)} = \lim_{{x \to 0^+}} \frac{1/x}{-1/x^2} = \lim_{{x \to 0^+}} -x = 0
\]
4. Since \(\lim_{{x \to 0^+}} \ln y = 0\), then \(y = e^0 = 1\).
### Conclusion:
Therefore,
\[
\lim_{{x \to 0^+}} x^x = 1
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff91ef73-4530-48ba-be88-fd15c629792d%2F939fe57a-4df1-4c88-aac1-1fd7455392f9%2F3xsr3t_processed.png&w=3840&q=75)
Transcribed Image Text:**27. \(\lim_{{x \to 0^+}} x^x\)**
---
**Explanation:**
To compute the limit \(\lim_{{x \to 0^+}} x^x\), we need to consider the behavior of the function \(f(x) = x^x\) as \(x\) approaches \(0\) from the positive side.
### Approach:
1. Consider using the natural logarithm to simplify the expression:
\[
y = x^x \implies \ln y = \ln (x^x) = x \ln x
\]
2. Now, evaluate the limit of \(\ln y\) as \(x\) approaches 0 from the right:
\[
\lim_{{x \to 0^+}} \ln y = \lim_{{x \to 0^+}} (x \ln x)
\]
3. Use L'Hôpital's Rule to evaluate this limit:
\[
\lim_{{x \to 0^+}} x \ln x \text{ is of the form } 0 \times -\infty, \text{ which we can rewrite as } \frac{\ln x}{\frac{1}{x}}
\]
Applying L'Hôpital's Rule:
\[
\lim_{{x \to 0^+}} \frac{\ln x}{1/x} = \lim_{{x \to 0^+}} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(1/x)} = \lim_{{x \to 0^+}} \frac{1/x}{-1/x^2} = \lim_{{x \to 0^+}} -x = 0
\]
4. Since \(\lim_{{x \to 0^+}} \ln y = 0\), then \(y = e^0 = 1\).
### Conclusion:
Therefore,
\[
\lim_{{x \to 0^+}} x^x = 1
\]
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