Math Problem Statement

Sketch the following function and use it to determine the values of a (list in ascending order below) for which lim f(x) does not exist: f(x) = {2 - x if x < -1, -x if -1 <= x < 1, (x - 1)^2 if x >= 1}

Solution

The image you've uploaded shows a problem related to the limits and piecewise functions. Here's the function from the image:

\begin{cases} 2 - x & \text{if } x < -1 \\ -x & \text{if } -1 \leq x < 1 \\ (x - 1)^2 & \text{if } x \geq 1 \end{cases}$$ The task is to determine the values of $$a$$ for which the limit of $$f(x)$$ does not exist as $$x \to a$$. To solve this: 1. **Sketch the function** to observe the behavior at the points where the pieces of the function change. These critical points are at $$x = -1$$ and $$x = 1$$, where the function transitions from one expression to another. 2. **Check the limits** from both sides at these points: - As $$x \to -1^-$$ (from the left) and $$x \to -1^+$$ (from the right). - As $$x \to 1^-$$ (from the left) and $$x \to 1^+$$ (from the right). 3. If the limits do not match, the function has no limit at that point. Let's go over these step by step: - At $$x = -1$$: - Left-hand limit: $$\lim_{x \to -1^-} (2 - x) = 2 - (-1) = 3$$ - Right-hand limit: $$\lim_{x \to -1^+} (-x) = -(-1) = 1$$ Since these limits do not match, the limit at $$x = -1$$ does not exist. - At $$x = 1$$: - Left-hand limit: $$\lim_{x \to 1^-} (-x) = -1$$ - Right-hand limit: $$\lim_{x \to 1^+} (x - 1)^2 = (1 - 1)^2 = 0$$ Since these limits do not match, the limit at $$x = 1$$ does not exist. ### Conclusion: The values of $$a$$ where the limit does not exist are $$a = -1$$ and $$a = 1$$. Would you like further clarification or details?

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Math Problem Analysis

Mathematical Concepts

Limits
Piecewise Functions
Continuity

Formulas

lim f(x) as x -> a
Piecewise Function Definition

Theorems

Limit Theorem for Piecewise Functions
Left-Hand and Right-Hand Limits

Suitable Grade Level

Grades 11-12