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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Prameter k
![---
### Evaluating Limits with Piecewise Functions
#### Given Function:
\[
f(x) =
\begin{cases}
2k - x & \text{if } x \geq 1 \\
\dfrac{1}{k} + x & \text{if } x < 1
\end{cases}
\]
Where \( k \) is a parameter.
#### Problem Statement:
Evaluate \(\lim_{{x \to 1^-}} f(x)\)
#### Question:
Can you choose a proper value for \( k \) so that \(\lim_{{x \to 1}} f(x)\) exists?
---
In this example, you are presented with a piecewise function \( f(x) \) that is defined differently for values of \( x \) greater than or equal to 1 and values of \( x \) less than 1. Your task is to evaluate the left-hand limit (\( \lim_{{x \to 1^-}} \)) of the function and determine if there is a value of the parameter \( k \) that makes the overall limit as \( x \) approaches 1 exist.
Let's break down the function:
- For \( x \geq 1 \), the function is \( f(x) = 2k - x \).
- For \( x < 1 \), the function is \( f(x) = \frac{1}{k} + x \).
To find if the proper \( k \) makes \(\lim_{{x \to 1}} f(x)\) exist, we need to evaluate the limits from both sides and see if they are equal:
1. **Evaluate the left-hand limit (\( \lim_{{x \to 1^-}} \)):**
\[ \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} \left( \frac{1}{k} + x \right) = \frac{1}{k} + 1 \]
2. **Evaluate the right-hand limit (\( \lim_{{x \to 1^+}} \)):**
\[ \lim_{{x \to 1}} f(x) = \lim_{{x \to 1^+}} \left( 2k - x \right) = 2k - 1 \]
3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc716be-e066-4234-bd1a-1a86837575c3%2F6bb53984-2bb8-47a9-be91-650f23967c36%2Fkxeh0tb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Evaluating Limits with Piecewise Functions
#### Given Function:
\[
f(x) =
\begin{cases}
2k - x & \text{if } x \geq 1 \\
\dfrac{1}{k} + x & \text{if } x < 1
\end{cases}
\]
Where \( k \) is a parameter.
#### Problem Statement:
Evaluate \(\lim_{{x \to 1^-}} f(x)\)
#### Question:
Can you choose a proper value for \( k \) so that \(\lim_{{x \to 1}} f(x)\) exists?
---
In this example, you are presented with a piecewise function \( f(x) \) that is defined differently for values of \( x \) greater than or equal to 1 and values of \( x \) less than 1. Your task is to evaluate the left-hand limit (\( \lim_{{x \to 1^-}} \)) of the function and determine if there is a value of the parameter \( k \) that makes the overall limit as \( x \) approaches 1 exist.
Let's break down the function:
- For \( x \geq 1 \), the function is \( f(x) = 2k - x \).
- For \( x < 1 \), the function is \( f(x) = \frac{1}{k} + x \).
To find if the proper \( k \) makes \(\lim_{{x \to 1}} f(x)\) exist, we need to evaluate the limits from both sides and see if they are equal:
1. **Evaluate the left-hand limit (\( \lim_{{x \to 1^-}} \)):**
\[ \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} \left( \frac{1}{k} + x \right) = \frac{1}{k} + 1 \]
2. **Evaluate the right-hand limit (\( \lim_{{x \to 1^+}} \)):**
\[ \lim_{{x \to 1}} f(x) = \lim_{{x \to 1^+}} \left( 2k - x \right) = 2k - 1 \]
3.
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