Math Problem Statement
Determine the values of a for which the limit as x approaches a of f(x) exists. The piecewise function f(x) is defined as:
1 + x if x < -1,
x^2 if -1 ≤ x < 1,
2 - x if x ≥ 1.
Solution
The given piecewise function is defined as:
\begin{cases} 1 + x & \text{if } x < -1, \\ x^2 & \text{if } -1 \leq x < 1, \\ 2 - x & \text{if } x \geq 1. \end{cases}$$ We are asked to find the values of $$a$$ for which $$\lim_{x \to a} f(x)$$ exists. This requires the left-hand limit and right-hand limit of $$f(x)$$ to be equal at any point $$a$$. ### Step-by-Step Analysis 1. **For $$a < -1$$:** Since the function $$f(x) = 1 + x$$ is the only expression defined for $$x < -1$$, it is continuous and the limit exists for all $$a < -1$$. Hence, $$\lim_{x \to a} f(x) = 1 + a$$ exists for any $$a < -1$$. 2. **At $$a = -1$$:** - **Left-hand limit:** As $$x \to -1^-$$, we use $$f(x) = 1 + x$$, so: $$\lim_{x \to -1^-} f(x) = 1 + (-1) = 0.$$ - **Right-hand limit:** As $$x \to -1^+$$, we use $$f(x) = x^2$$, so: $$\lim_{x \to -1^+} f(x) = (-1)^2 = 1.$$ Since $$\lim_{x \to -1^-} f(x) = 0$$ and $$\lim_{x \to -1^+} f(x) = 1$$, the limit does not exist at $$a = -1$$. 3. **For $$-1 < a < 1$$:** In this interval, $$f(x) = x^2$$, which is a continuous function. Therefore, the limit exists for any $$a$$ in the interval $$-1 < a < 1$$, and $$\lim_{x \to a} f(x) = a^2$$. 4. **At $$a = 1$$:** - **Left-hand limit:** As $$x \to 1^-$$, we use $$f(x) = x^2$$, so: $$\lim_{x \to 1^-} f(x) = 1^2 = 1.$$ - **Right-hand limit:** As $$x \to 1^+$$, we use $$f(x) = 2 - x$$, so: $$\lim_{x \to 1^+} f(x) = 2 - 1 = 1.$$ Since $$\lim_{x \to 1^-} f(x) = 1$$ and $$\lim_{x \to 1^+} f(x) = 1$$, the limit exists at $$a = 1$$, and $$\lim_{x \to 1} f(x) = 1$$. 5. **For $$a > 1$$:** Since $$f(x) = 2 - x$$ is the only expression defined for $$x > 1$$, the function is continuous and the limit exists for all $$a > 1$$. Hence, $$\lim_{x \to a} f(x) = 2 - a$$ exists for any $$a > 1$$. ### Conclusion: The limit $$\lim_{x \to a} f(x)$$ exists for all $$a$$ except $$a = -1$$. Therefore, the values of $$a$$ for which the limit exists are: $$a \in (-\infty, -1) \cup (-1, 1] \cup (1, \infty).$$ Would you like more details on any step? ### Related Questions: 1. How is the definition of continuity used to verify the existence of limits? 2. What is the significance of one-sided limits in piecewise functions? 3. How can we extend this concept to higher-dimensional piecewise functions? 4. What role does differentiability play in the behavior of limits at the points of discontinuity? 5. Can you show an example where a limit exists but the function is not continuous at that point? **Tip:** When evaluating limits at points where a piecewise function changes definition, always consider both the left-hand and right-hand limits.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Continuity
Formulas
Limit of a function
Left-hand limit
Right-hand limit
Theorems
Continuity Theorem
Limit Definition
Suitable Grade Level
Grades 11-12
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