Math Problem Statement
Với điều kiện nào thì hàm số f(x) = { x^n sin(1/x) nếu x ≠ 0, 0 nếu x = 0 } (n ∈ Z) liên tục, khả vi, có đạo hàm liên tục tại x = 0?
Solution
The given function is defined as:
x^n \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$$ with $$n \in \mathbb{Z}$$ (integer). The problem asks under what conditions the function $$f(x)$$ satisfies the following: 1. **Continuity at $$x = 0$$**. 2. **Differentiability at $$x = 0$$**. 3. **Continuously differentiable at $$x = 0$$**. ### Let's analyze each condition: ### a) Continuity at $$x = 0$$: For $$f(x)$$ to be continuous at $$x = 0$$, we must have: $$\lim_{x \to 0} f(x) = f(0) = 0.$$ Now, examine the limit: $$\lim_{x \to 0} f(x) = \lim_{x \to 0} x^n \sin\left(\frac{1}{x}\right).$$ Since $$\sin\left(\frac{1}{x}\right)$$ is bounded between -1 and 1, the limit behavior of $$f(x)$$ depends on the factor $$x^n$$. - If $$n > 0$$, as $$x^n \to 0$$ when $$x \to 0$$, the whole expression tends to 0. - If $$n \leq 0$$, the limit does not tend to 0 because the oscillations of $$\sin\left(\frac{1}{x}\right)$$ dominate. Therefore, the function is continuous at $$x = 0$$ if and only if $$n > 0$$. ### b) Differentiability at $$x = 0$$: For differentiability at $$x = 0$$, the derivative: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{x^n \sin\left(\frac{1}{x}\right)}{h}$$ must exist. The term $$x^n \sin\left(\frac{1}{x}\right)$$ is tricky because of the oscillatory nature of the sine function. By a more rigorous analysis using bounds on $$\sin\left(\frac{1}{x}\right)$$ and the behavior of $$x^n$$, it can be shown that $$f(x)$$ is differentiable if $$n \geq 1$$. ### c) Continuously differentiable at $$x = 0$$: For the derivative to be continuous at $$x = 0$$, we need to check if: $$\lim_{x \to 0} f'(x) = f'(0).$$ The condition for continuously differentiable is stricter, and typically the analysis shows that this holds when $$n \geq 2$$. ### Summary of conditions: - **Continuity at $$x = 0$$**: $$n > 0$$. - **Differentiability at $$x = 0$$**: $$n \geq 1$$. - **Continuously differentiable at $$x = 0$$**: $$n \geq 2$$. Would you like further details or have any specific questions? Here are five related questions for further exploration: 1. How do we rigorously prove the limits involved in continuity and differentiability? 2. What happens when $$n = 0$$? Why does the function become discontinuous? 3. How does the oscillatory behavior of $$\sin\left(\frac{1}{x}\right)$$ affect differentiability? 4. Can we generalize this result for other oscillating functions like $$\cos\left(\frac{1}{x}\right)$$? 5. What is the physical significance of requiring continuous derivatives in applications? **Tip**: When analyzing limits of oscillating functions, it’s helpful to remember that bounded oscillations can often simplify to focusing on the power of $$x$$.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Continuity
Differentiability
Continuous Differentiability
Limits
Formulas
f(x) = x^n sin(1/x) if x ≠ 0, 0 if x = 0
Continuity: lim_{x -> 0} f(x) = f(0)
Differentiability: lim_{h -> 0} (f(h) - f(0)) / h
Continuous differentiability: lim_{x -> 0} f'(x) = f'(0)
Theorems
Limit Theorem
Squeeze Theorem
Definition of Continuity
Definition of Differentiability
Suitable Grade Level
University (Calculus)
Related Recommendation
Differentiability of f(x) = x^2 sin(sin(1/x)) at x = 0
Analyzing Differentiability of f(x) = x^2 sin(sin(x^-1)) at x = 0
Continuity of a Piecewise Function: sin(x)/x at x = 0
Limit and Continuity of Oscillatory Function with Discontinuity at x=0
Finding Continuity and Differentiability of a Piecewise Function at x=0