Math Problem Statement
If f is the function defined above, then \( \lim_{x \to 2} f(x) \) is
Solution
To solve the problem, we need to determine the limit of the function as approaches 2.
Given the piecewise function:
\frac{(x-2)^2 \sin x}{|x-2|} & \text{for } x \neq 2 \\ \pi & \text{for } x = 2 \end{cases}$$ We are asked to find $$\lim_{x \to 2} f(x)$$. ### Step 1: Evaluate the limit from the left and right as $$x$$ approaches 2 Since $$f(x)$$ is defined differently for $$x = 2$$, we first focus on the part of the function where $$x \neq 2$$: $$f(x) = \frac{(x-2)^2 \sin x}{|x-2|}$$ For $$x \neq 2$$, let's evaluate the limit as $$x$$ approaches 2: 1. **Case 1: $$x > 2$$ (Right-hand limit)** When $$x > 2$$, $$|x - 2| = x - 2$$. Therefore, the function simplifies to: $$f(x) = \frac{(x-2)^2 \sin x}{x-2} = (x-2)\sin x$$ Now, taking the limit as $$x$$ approaches 2: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x-2)\sin x$$ Since $$x - 2$$ approaches 0 as $$x$$ approaches 2, and $$\sin x$$ is bounded, this limit is: $$\lim_{x \to 2^+} f(x) = 0$$ 2. **Case 2: $$x < 2$$ (Left-hand limit)** When $$x < 2$$, $$|x - 2| = 2 - x$$. Therefore, the function simplifies to: $$f(x) = \frac{(x-2)^2 \sin x}{-(x-2)} = -(x-2)\sin x$$ Taking the limit as $$x$$ approaches 2: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} -(x-2)\sin x$$ Again, since $$x - 2$$ approaches 0, the limit is: $$\lim_{x \to 2^-} f(x) = 0$$ ### Step 2: Compare the left-hand and right-hand limits Both the left-hand limit $$\lim_{x \to 2^-} f(x)$$ and the right-hand limit $$\lim_{x \to 2^+} f(x)$$ are equal to 0. ### Conclusion Since the left-hand limit and right-hand limit are equal, the limit exists and is equal to 0. Thus, $$\lim_{x \to 2} f(x) = 0$$, so the correct answer is **(A) 0**. Would you like further details on any part of this problem? Here are five related questions that could expand your understanding: 1. What is the significance of left-hand and right-hand limits in determining the overall limit? 2. How would the limit change if the function had a different form for $$x = 2$$? 3. What are some common scenarios where a limit does not exist? 4. How does the behavior of $$\sin x$$ near $$x = 2$$ influence the result? 5. Can the value of a function at a point affect the limit at that point? **Tip:** When evaluating limits, always check both the left-hand and right-hand limits, especially when dealing with piecewise functions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Trigonometry
Formulas
\( \lim_{x \to c} f(x) \)
Sine function: \( \sin(x) \)
Theorems
Limit of a function at a point
Suitable Grade Level
Grade 11-12