Math Problem Statement
Find the requested limits for the function f(x) defined as: f(x) = 2 - 2x if x < 2, x ≠ 1; f(x) = -1 / (x - 4) if x > 2, x ≠ 7. Also, check if f is continuous at x = 7 and if any discontinuity is reparable.
Solution
To solve the problem, I'll analyze each part step-by-step. The function is defined as follows:
2 - 2x & \text{if } x < 2, \, x \neq 1 \\ \frac{-1}{x - 4} & \text{if } x > 2, \, x \neq 7 \end{cases}$$ Let's address each limit and the continuity question: ### a. $$\lim_{x \to 1} f(x)$$ Since $$x = 1$$ falls under the condition $$x < 2$$, we use the first piece of the function: $$f(x) = 2 - 2x$$. $$\lim_{x \to 1} (2 - 2x) = 2 - 2(1) = 0$$ Thus, $$\lim_{x \to 1} f(x) = 0$$. ### b. $$\lim_{x \to 2^-} f(x)$$ Approaching $$x = 2$$ from the left, we use the first piece of the function $$f(x) = 2 - 2x$$: $$\lim_{x \to 2^-} (2 - 2x) = 2 - 2(2) = -2$$ So, $$\lim_{x \to 2^-} f(x) = -2$$. ### c. $$\lim_{x \to 2^+} f(x)$$ Approaching $$x = 2$$ from the right, we use the second piece of the function $$f(x) = \frac{-1}{x - 4}$$: $$\lim_{x \to 2^+} \frac{-1}{x - 4} = \frac{-1}{2 - 4} = \frac{-1}{-2} = \frac{1}{2}$$ Thus, $$\lim_{x \to 2^+} f(x) = \frac{1}{2}$$. ### d. $$\lim_{x \to 7} f(x)$$ Since $$x = 7$$ falls under $$x > 2$$, we use the second piece: $$f(x) = \frac{-1}{x - 4}$$: $$\lim_{x \to 7} \frac{-1}{x - 4} = \frac{-1}{7 - 4} = \frac{-1}{3}$$ Therefore, $$\lim_{x \to 7} f(x) = \frac{-1}{3}$$. ### e. Is $$f$$ continuous at $$x = 7$$? If not, is this reparable? To check continuity at $$x = 7$$, the function must satisfy: 1. $$f(7)$$ is defined, 2. $$\lim_{x \to 7} f(x)$$ exists, 3. $$\lim_{x \to 7} f(x) = f(7)$$. The function $$f$$ is not defined at $$x = 7$$, so $$f(7)$$ does not exist. Therefore, $$f$$ is not continuous at $$x = 7$$. **Reparability**: The discontinuity is removable. If we define $$f(7) = \frac{-1}{3}$$, then the function would become continuous at $$x = 7$$. ### Summary of Results: 1. $$\lim_{x \to 1} f(x) = 0$$ 2. $$\lim_{x \to 2^-} f(x) = -2$$ 3. $$\lim_{x \to 2^+} f(x) = \frac{1}{2}$$ 4. $$\lim_{x \to 7} f(x) = \frac{-1}{3}$$ 5. $$f$$ is not continuous at $$x = 7$$, but the discontinuity is reparable by defining $$f(7) = \frac{-1}{3}$$. Would you like further details or have any questions? Here are some follow-up questions to consider: 1. What does it mean for a function to be piecewise continuous? 2. How would you determine if a limit exists when $$x$$ approaches a specific point? 3. Can a function have a limit at a point where it is not defined? 4. What is the difference between a removable and a non-removable discontinuity? 5. How can a function be modified to make it continuous at a point of removable discontinuity? **Tip**: When determining continuity at a point, always check if the limit from the left matches the limit from the right, and if both equal the function's value at that point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Piecewise Functions
Formulas
Limit Definition
Continuity Condition
Theorems
Definition of Limits
Removable Discontinuities
Suitable Grade Level
Undergraduate Calculus
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