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The problem asks to check the differentiability of the function \( f(x) \) given by:

\[
f(x) = 
\begin{cases} 
x^2 \cos\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\
0, & \text{if } x = 0 
\end{cases}
\]

at \( x = 0 \) and \( x = \frac{1}{\pi} \).

### Step-by-Step Approach:

#### 1. Differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to check:
- Continuity at \( x = 0 \)
- Existence of the derivative \( f'(0) \)

**Continuity at \( x = 0 \):**
We check whether \( \lim_{x \to 0} f(x) = f(0) \).

Given \( f(0) = 0 \), we now find:

\[
\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)
\]

Since \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1 for all \( x \), and since \( x^2 \) approaches 0 as \( x \to 0 \), we have:

\[
- x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2
\]

By the squeeze theorem:

\[
\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0
\]

Thus, \( \lim_{x \to 0} f(x) = 0 = f(0) \), so \( f(x) \) is continuous at \( x = 0 \).

**Existence of derivative at \( x = 0 \):**

We need to check if the derivative \( f'(0) \) exists, defined as:

\[
f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}
\]

Substitute \( f(0) = 0 \):

\[
f'(0) = \lim_{h \to 0} \frac{h^2 \cos\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \cos\left(\frac{1}{h}\right)
\]

Since \( \cos\left(\frac{1}{h}\right) \) oscillates between -1 and 1, and \( h \to 0 \), by the squeeze theorem:

\[
\lim_{h \to 0} h \cos\left(\frac{1}{h}\right) = 0
\]

Thus, \( f'(0) = 0 \), and \( f(x) \) is differentiable at \( x = 0 \).

#### 2. Differentiability at \( x = \frac{1}{\pi} \)

Since \( x \neq 0 \), we can differentiate the function directly. For \( f(x) = x^2 \cos\left(\frac{1}{x}\right) \), we apply the product rule:

\[
f'(x) = \frac{d}{dx}\left(x^2 \cos\left(\frac{1}{x}\right)\right) = 2x \cos\left(\frac{1}{x}\right) - \sin\left(\frac{1}{x}\right)
\]

Now, substitute \( x = \frac{1}{\pi} \):

\[
f'\left(\frac{1}{\pi}\right) = 2 \left(\frac{1}{\pi}\right) \cos(\pi) - \sin(\pi)
\]

Since \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \), we have:

\[
f'\left(\frac{1}{\pi}\right) = 2 \left(\frac{1}{\pi}\right)(-1) = -\frac{2}{\pi}
\]

Thus, \( f(x) \) is differentiable at \( x = \frac{1}{\pi} \), and the derivative is \( f'\left(\frac{1}{\pi}\right) = -\frac{2}{\pi} \).

---

Let me know if you'd like more details! Here are 5 related questions to further explore the concepts:

1. What are the conditions for differentiability of a piecewise function?
2. How does the squeeze theorem work in proving limits?
3. Can a function be continuous but not differentiable at a point?
4. What role do oscillating functions like \( \cos\left(\frac{1}{x}\right) \) play in limiting behavior?
5. How do product and chain rules combine when differentiating products of functions?

**Tip:** When checking differentiability, always ensure that both continuity and the existence of the derivative are considered.

Check the differentiability of the function f(x) = { x^2 * cos(1/x) if x ≠ 0, 0 if x = 0 } at x = 0 and x = 1/π.

This problem explores the differentiability of the piecewise function f(x) = x^2 * cos(1/x) for x ≠ 0 and 0 for x = 0, at x = 0 and x = 1/π. The solution covers limits, the squeeze theorem, and the definition of differentiability.

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