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...
Related questions
Question
I need help with this homework problem please provide steps and details
on how you got the answer so I can learn and compare my work
![### Example Problem: Evaluating a Limit
**Evaluate the given limit, and justify each step by indicating the limit law(s) used.**
\[ \lim_{{x \to -2}} \frac{\sqrt{x^2 + 5} - 3}{x + 2} \]
---
1. **Identify the Form of the Limit:**
- First, substitute \( x = -2 \) into the expression to check the form:
\[
\frac{\sqrt{(-2)^2 + 5} - 3}{-2 + 2} = \frac{\sqrt{4 + 5} - 3}{0} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}
\]
- Since the limit results in the indeterminate form \(\frac{0}{0}\), we need further simplification.
2. **Simplify the Expression:**
- To simplify, multiply the numerator and the denominator by the conjugate of the numerator:
\[
\frac{\sqrt{x^2 + 5} - 3}{x + 2} \cdot \frac{\sqrt{x^2 + 5} + 3}{\sqrt{x^2 + 5} + 3} = \frac{(\sqrt{x^2 + 5} - 3)(\sqrt{x^2 + 5} + 3)}{(x + 2)(\sqrt{x^2 + 5} + 3)}
\]
- The numerator simplifies using the difference of squares:
\[
\frac{(x^2 + 5) - 9}{(x + 2)(\sqrt{x^2 + 5} + 3)} = \frac{x^2 - 4}{(x + 2)(\sqrt{x^2 + 5} + 3)}
\]
- The numerator can be factored further:
\[
x^2 - 4 = (x - 2)(x + 2)
\]
- Substitute the factored numerator:
\[
\frac{(x - 2)(x + 2)}{(x + 2)(\sqrt{x^2 + 5} + 3)}
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff62d5ca7-5439-4cfe-bb83-29d1381eaedb%2F9b6ff0aa-2107-49f3-a240-ea666a77e257%2Fwjjjqr8s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem: Evaluating a Limit
**Evaluate the given limit, and justify each step by indicating the limit law(s) used.**
\[ \lim_{{x \to -2}} \frac{\sqrt{x^2 + 5} - 3}{x + 2} \]
---
1. **Identify the Form of the Limit:**
- First, substitute \( x = -2 \) into the expression to check the form:
\[
\frac{\sqrt{(-2)^2 + 5} - 3}{-2 + 2} = \frac{\sqrt{4 + 5} - 3}{0} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}
\]
- Since the limit results in the indeterminate form \(\frac{0}{0}\), we need further simplification.
2. **Simplify the Expression:**
- To simplify, multiply the numerator and the denominator by the conjugate of the numerator:
\[
\frac{\sqrt{x^2 + 5} - 3}{x + 2} \cdot \frac{\sqrt{x^2 + 5} + 3}{\sqrt{x^2 + 5} + 3} = \frac{(\sqrt{x^2 + 5} - 3)(\sqrt{x^2 + 5} + 3)}{(x + 2)(\sqrt{x^2 + 5} + 3)}
\]
- The numerator simplifies using the difference of squares:
\[
\frac{(x^2 + 5) - 9}{(x + 2)(\sqrt{x^2 + 5} + 3)} = \frac{x^2 - 4}{(x + 2)(\sqrt{x^2 + 5} + 3)}
\]
- The numerator can be factored further:
\[
x^2 - 4 = (x - 2)(x + 2)
\]
- Substitute the factored numerator:
\[
\frac{(x - 2)(x + 2)}{(x + 2)(\sqrt{x^2 + 5} + 3)}
\
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning