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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![### Limits of Piecewise Functions and Rational Functions
#### Problem Statement:
Evaluate the following limits:
e. \[
\lim_{{x \to -2^+}} f(x), \quad \text{where} \quad f(x) =
\begin{cases}
5x + 1 & \text{for} \quad x > -2 \\
x^2 & \text{for} \quad x \leq -2
\end{cases}
\]
f. \[
\lim_{{x \to -1}} \frac{x + 2}{x + 1}
\]
#### Explanation:
**Problem e: Limit of a piecewise function from the right**
- For \(x > -2\), the function \(f(x)\) is defined as \(5x + 1\).
- For \(x \leq -2\), the function \(f(x)\) is defined as \(x^2\).
We are asked to find \(\lim_{{x \to -2^+}} f(x)\), which means we need to evaluate the limit of \(f(x)\) as \(x\) approaches \(-2\) from the right (values of \(x\) slightly greater than \(-2\)).
**Problem f: Limit of a rational function**
- This is a limit of a rational function where the numerator is \(x + 2\) and the denominator is \(x + 1\), evaluated as \(x\) approaches \(-1\).
### Detailed Steps to Solve:
**e. Evaluating \(\lim_{{x \to -2^+}} f(x)\)**
Since \(x \to -2^+\) implies approaching \(-2\) from the right, we use the part of the piecewise function that applies to \(x > -2\):
For \(x > -2\):
\[ f(x) = 5x + 1 \]
So, we need:
\[ \lim_{x \to -2^+} (5x + 1) \]
When \(x\) approaches \(-2\), we substitute \( -2 \) into \(5x + 1\):
\[ 5(-2) + 1 = -10 + 1 = -9 \]
Therefore:
\[ \lim_{x \to -2^+} f(x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc59faf40-90fa-4a45-a3f2-1ae0c0471a65%2Fe9a8cd00-cfb5-433a-8c03-b77fcd38bf0b%2Fhpyxh5.png&w=3840&q=75)
Transcribed Image Text:### Limits of Piecewise Functions and Rational Functions
#### Problem Statement:
Evaluate the following limits:
e. \[
\lim_{{x \to -2^+}} f(x), \quad \text{where} \quad f(x) =
\begin{cases}
5x + 1 & \text{for} \quad x > -2 \\
x^2 & \text{for} \quad x \leq -2
\end{cases}
\]
f. \[
\lim_{{x \to -1}} \frac{x + 2}{x + 1}
\]
#### Explanation:
**Problem e: Limit of a piecewise function from the right**
- For \(x > -2\), the function \(f(x)\) is defined as \(5x + 1\).
- For \(x \leq -2\), the function \(f(x)\) is defined as \(x^2\).
We are asked to find \(\lim_{{x \to -2^+}} f(x)\), which means we need to evaluate the limit of \(f(x)\) as \(x\) approaches \(-2\) from the right (values of \(x\) slightly greater than \(-2\)).
**Problem f: Limit of a rational function**
- This is a limit of a rational function where the numerator is \(x + 2\) and the denominator is \(x + 1\), evaluated as \(x\) approaches \(-1\).
### Detailed Steps to Solve:
**e. Evaluating \(\lim_{{x \to -2^+}} f(x)\)**
Since \(x \to -2^+\) implies approaching \(-2\) from the right, we use the part of the piecewise function that applies to \(x > -2\):
For \(x > -2\):
\[ f(x) = 5x + 1 \]
So, we need:
\[ \lim_{x \to -2^+} (5x + 1) \]
When \(x\) approaches \(-2\), we substitute \( -2 \) into \(5x + 1\):
\[ 5(-2) + 1 = -10 + 1 = -9 \]
Therefore:
\[ \lim_{x \to -2^+} f(x
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