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The given function is defined as:

\[
f(x) = \begin{cases} 
x^n \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\
0 & \text{if } x = 0 
\end{cases}
\]
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 \(x\).

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?

Explore the conditions under which the function f(x) = x^n sin(1/x) (if x ≠ 0) and f(x) = 0 (if x = 0) is continuous, differentiable, and continuously differentiable at x = 0. This problem involves limits, continuity, and differentiability concepts, suitable for advanced calculus studies.

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