Math Problem Statement
Evaluate the following limits of the piecewise function f(x) as x approaches -1 and 1 from both sides.
Solution
We are given a piecewise function and need to evaluate limits as and . Let's analyze the function and the limits.
Function Definition:
\begin{cases} x + 2, & \text{if } x \leq -1 \\ -x^2 + 2x, & \text{if } -1 < x < 1 \\ \frac{1}{x}, & \text{if } x > 1 \end{cases}$$ ### Limits to evaluate: - $$\lim_{x \to -1^-} f(x)$$ - $$\lim_{x \to -1^+} f(x)$$ - $$\lim_{x \to 1^-} f(x)$$ - $$\lim_{x \to 1^+} f(x)$$ ### a. $$\lim_{x \to -1^-} f(x)$$: For $$x \to -1^-$$ (from the left side of $$-1$$), the relevant piece of the function is $$f(x) = x + 2$$ (since $$x \leq -1$$). $$\lim_{x \to -1^-} f(x) = (-1) + 2 = 1$$ ### b. $$\lim_{x \to -1^+} f(x)$$: For $$x \to -1^+$$ (from the right side of $$-1$$), the relevant piece is $$f(x) = -x^2 + 2x$$ (since $$-1 < x < 1$$). $$\lim_{x \to -1^+} f(x) = -(-1)^2 + 2(-1) = -1 - 2 = -3$$ ### c. $$\lim_{x \to 1^-} f(x)$$: For $$x \to 1^-$$ (from the left side of $$1$$), the relevant piece is $$f(x) = -x^2 + 2x$$ (since $$-1 < x < 1$$). $$\lim_{x \to 1^-} f(x) = -(1)^2 + 2(1) = -1 + 2 = 1$$ ### d. $$\lim_{x \to 1^+} f(x)$$: For $$x \to 1^+$$ (from the right side of $$1$$), the relevant piece is $$f(x) = \frac{1}{x}$$ (since $$x > 1$$). $$\lim_{x \to 1^+} f(x) = \frac{1}{1} = 1$$ ### Summary of Limits: - $$\lim_{x \to -1^-} f(x) = 1$$ - $$\lim_{x \to -1^+} f(x) = -3$$ - $$\lim_{x \to 1^-} f(x) = 1$$ - $$\lim_{x \to 1^+} f(x) = 1$$ Do you need more details or have any questions? ### Related Questions: 1. How do you determine if a function is continuous at a given point? 2. What are the conditions for a limit to exist at a given point? 3. What is the significance of the left-hand and right-hand limits? 4. How would the function's behavior change if the piecewise conditions were modified? 5. Can you generalize this method for finding limits of other piecewise functions? ### Tip: When evaluating limits of piecewise functions, always check which piece applies in the neighborhood of the point from both sides (left and right).Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Continuity
Formulas
Limit from the left: lim(x → a⁻) f(x)
Limit from the right: lim(x → a⁺) f(x)
Theorems
Left-hand limit and right-hand limit must be equal for the limit to exist at a point.
Suitable Grade Level
Grades 10-12