Math Problem Statement
Consider the function f(x) = {1/x^3 * sin(1/x^3), x ≠ 0; 0, x = 0}. Then which of the following is not true?
Solution
Let's analyze the given problem.
The function is defined as:
\begin{cases} \frac{1}{x^3} \sin \left( \frac{1}{x^3} \right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$$ We are asked to identify which statement is **not true** from the following options: a. None of these b. The function is discontinuous at $$x = 0$$ c. The limit exists at $$x = 0$$ d. The limit does not exist at $$x = 0$$ ### Step-by-Step Analysis: 1. **Limit behavior as $$x \to 0$$:** - For $$x \neq 0$$, the function is $$\frac{1}{x^3} \sin \left( \frac{1}{x^3} \right)$$. We know that $$\sin(\theta)$$ is bounded between $$-1$$ and $$1$$, so: $$-\frac{1}{x^3} \leq \frac{1}{x^3} \sin \left( \frac{1}{x^3} \right) \leq \frac{1}{x^3}$$ As $$x \to 0$$, the oscillations of $$\sin \left( \frac{1}{x^3} \right)$$ become wilder, leading the function to oscillate more erratically without settling on a single value. Therefore, **the limit does not exist** as $$x \to 0$$. 2. **Discontinuity at $$x = 0$$:** - The function is given as 0 at $$x = 0$$. However, as shown in the previous step, the limit of the function as $$x \to 0$$ does not exist. Therefore, the function is **discontinuous at $$x = 0$$**. 3. **Analyzing the options:** - **Option (b)**: The function is indeed discontinuous at $$x = 0$$. This statement is **true**. - **Option (c)**: The limit does not exist at $$x = 0$$. Therefore, this statement is **false**. - **Option (d)**: The limit does not exist, so this statement is **true**. - **Option (a)**: Since option (c) is false, this statement is also **false**. ### Conclusion: The correct answer is **(c)**, as the statement "The limit exists at $$x = 0$$" is **not true**. Would you like further explanation on any part of the solution? Here are some related questions to deepen your understanding: 1. How can you formally prove that the limit does not exist for this function at $$x = 0$$? 2. What are some common behaviors of oscillating functions as $$x \to 0$$? 3. How do we test for discontinuity in functions like this one? 4. Why does the value of $$\sin \left( \frac{1}{x^3} \right)$$ cause such erratic behavior as $$x \to 0$$? 5. What are other examples of functions where the limit does not exist at a particular point due to oscillation? **Tip:** Always check both one-sided limits when determining whether a limit exists at a point. If the one-sided limits differ, the overall limit does not exist.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Oscillatory Functions
Formulas
Limit of a function as x approaches 0
Squeeze theorem for oscillatory functions
Theorems
Discontinuity at a point
Non-existence of a limit due to oscillation
Suitable Grade Level
Grades 11-12 (Pre-Calculus/Calculus)
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