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We are given the function \( f(x) = x^2 \sin\left( \sin\left( \frac{1}{x} \right) \right) \) for \( x \neq 0 \) and \( f(0) = 0 \). We are asked which of the following is true:

(A) \( f(x) \) is differentiable everywhere.

(B) \( f'(0) \neq 0 \).

(C) \( f'(0) = 0 \).

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### 1. **Behavior of the Function at \( x = 0 \):**

First, let's consider the behavior of \( f(x) \) around \( x = 0 \). The term \( \frac{1}{x} \) causes potential oscillations in the sine function as \( x \) approaches zero. Therefore, we need to check whether the function \( f(x) \) behaves smoothly and whether it is differentiable at \( x = 0 \).

### 2. **Limit of \( f(x) \) as \( x \to 0 \):**

When \( x = 0 \), we are given that \( f(0) = 0 \). Now, we need to check whether the limit of \( f(x) \) as \( x \to 0 \) exists and is continuous.

- For small \( x \), \( \frac{1}{x} \) becomes large, and \( \sin\left( \frac{1}{x} \right) \) oscillates between \( -1 \) and \( 1 \). However, since \( \sin \left( \sin \left( \frac{1}{x} \right) \right) \) is bounded between \( -1 \) and \( 1 \), the term \( x^2 \sin\left( \sin\left( \frac{1}{x} \right) \right) \) approaches zero as \( x \to 0 \), because \( x^2 \) dominates the oscillations of the sine function.

Therefore, the limit of \( f(x) \) as \( x \to 0 \) is zero, and the function is continuous at \( x = 0 \).

### 3. **Differentiability at \( x = 0 \):**

To check whether \( f(x) \) is differentiable at \( x = 0 \), we need to compute \( f'(0) \) using the definition of the derivative:
\[
f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}
\]
Since \( f(0) = 0 \), this simplifies to:
\[
f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{h^2 \sin\left( \sin\left( \frac{1}{h} \right) \right)}{h} = \lim_{h \to 0} h \sin\left( \sin\left( \frac{1}{h} \right) \right)
\]
Again, since \( \sin\left( \sin\left( \frac{1}{h} \right) \right) \) is bounded between \( -1 \) and \( 1 \), this limit becomes:
\[
f'(0) = \lim_{h \to 0} h \cdot \text{(bounded term)} = 0
\]
Thus, \( f'(0) = 0 \), meaning the function is differentiable at \( x = 0 \) and the derivative at 0 is zero.

### 4. **Differentiability Everywhere Else:**

For \( x \neq 0 \), the function \( f(x) = x^2 \sin\left( \sin\left( \frac{1}{x} \right) \right) \) is clearly differentiable because it's a composition of differentiable functions. The problem arises only at \( x = 0 \).

### Conclusion:

- (A) **True**: The function is differentiable everywhere, including at \( x = 0 \).
- (B) **False**: \( f'(0) = 0 \), so this statement is false.
- (C) **True**: \( f'(0) = 0 \), so this statement is true.

The correct answers are **(A)** and **(C)**.

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Do you have any questions, or would you like further clarification? Here are some related questions to explore:

1. What are the conditions for a function to be differentiable at a point?
2. How can you check the differentiability of a piecewise function at a boundary point?
3. Why do oscillating terms like \( \sin\left( \frac{1}{x} \right) \) cause trouble near zero?
4. How does L'Hopital's Rule help with limits involving oscillating functions?
5. What are some common techniques for finding derivatives of functions with oscillations?

**Tip:** When dealing with oscillating functions near singularities, always focus on the behavior of the dominant term, which can often simplify the analysis!

Which of the following is true for the function f(x) = x^2 sin(sin(1/x)) with f(0) = 0? Options: (A) differentiable everywhere, (B) f'(0) ≠ 0, (C) f'(0) = 0.

This problem explores the differentiability of the function f(x) = x^2 sin(sin(1/x)) with f(0) = 0. It covers concepts like oscillating functions, limits, and uses the Squeeze Theorem to find that f'(0) = 0 and that the function is differentiable everywhere.

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