D D ▷ TEXA fex) (2k- x 1 Fx k evaluate lim Can you choose Parameter k proper value for k x21 x < 1 a So lim fex) exists?

College Algebra
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ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter2: Functions
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Problem 30P: In this problem you are asked to find a function that models in real life situation and then use the...
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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.
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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