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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Find the one-sided limits
![### Understanding Piecewise Functions and Limits
In this exercise, we are given a piecewise function \( f(x) \) and asked to evaluate its limits as \( x \) approaches 3 from both the left and the right.
The piecewise function is defined as follows:
\[
f(x) =
\begin{cases}
x^2 + 2 & \text{if } 0 \leq x \leq 3 \\
2x + 5 & \text{if } x > 3
\end{cases}
\]
**Tasks:**
a. Calculate the left-hand limit of \( f(x) \) as \( x \) approaches 3:
\[ \lim_{x \to 3^-} f(x) \]
b. Calculate the right-hand limit of \( f(x) \) as \( x \) approaches 3:
\[ \lim_{x \to 3^+} f(x) \]
To solve these limits, observe the behavior of the functions defined for the intervals as \( x \) approaches 3 from the left and right:
1. **For \( x \to 3^- \):**
Since \( 0 \leq x \leq 3 \), the relevant part of the piecewise function is \( f(x) = x^2 + 2 \).
Thus,
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 + 2) \]
\[ = (3)^2 + 2 = 9 + 2 = 11 \]
2. **For \( x \to 3^+ \):**
Since \( x > 3 \), the relevant part of the piecewise function is \( f(x) = 2x + 5 \).
Thus,
\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x + 5) \]
\[ = 2(3) + 5 = 6 + 5 = 11 \]
### Conclusion:
Both the left-hand and right-hand limits as \( x \) approaches 3 are equal. Therefore, the limit of \( f(x) \) as \( x \) approaches 3 exists and is 11:
\[ \lim_{x \to 3} f(x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9e33d54-6f37-42c0-9ba8-1e66f868cdc7%2Fc4bdff07-ed1f-4076-a426-21e2aba6342d%2F15lln06_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Piecewise Functions and Limits
In this exercise, we are given a piecewise function \( f(x) \) and asked to evaluate its limits as \( x \) approaches 3 from both the left and the right.
The piecewise function is defined as follows:
\[
f(x) =
\begin{cases}
x^2 + 2 & \text{if } 0 \leq x \leq 3 \\
2x + 5 & \text{if } x > 3
\end{cases}
\]
**Tasks:**
a. Calculate the left-hand limit of \( f(x) \) as \( x \) approaches 3:
\[ \lim_{x \to 3^-} f(x) \]
b. Calculate the right-hand limit of \( f(x) \) as \( x \) approaches 3:
\[ \lim_{x \to 3^+} f(x) \]
To solve these limits, observe the behavior of the functions defined for the intervals as \( x \) approaches 3 from the left and right:
1. **For \( x \to 3^- \):**
Since \( 0 \leq x \leq 3 \), the relevant part of the piecewise function is \( f(x) = x^2 + 2 \).
Thus,
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 + 2) \]
\[ = (3)^2 + 2 = 9 + 2 = 11 \]
2. **For \( x \to 3^+ \):**
Since \( x > 3 \), the relevant part of the piecewise function is \( f(x) = 2x + 5 \).
Thus,
\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x + 5) \]
\[ = 2(3) + 5 = 6 + 5 = 11 \]
### Conclusion:
Both the left-hand and right-hand limits as \( x \) approaches 3 are equal. Therefore, the limit of \( f(x) \) as \( x \) approaches 3 exists and is 11:
\[ \lim_{x \to 3} f(x
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