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To determine if \( f'(0) \) exists for both functions \( f(x) \) in problems 59 and 60, we need to examine the differentiability at \( x = 0 \). We will analyze the definition of the derivative at \( x = 0 \), which is:

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

Since \( f(0) = 0 \) in both cases, this simplifies to:

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

Let's evaluate this limit for each function individually.

### Problem 59
The function is given by:

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

To find \( f'(0) \), we need to compute:

\[
f'(0) = \lim_{x \to 0} \frac{x \sin \left( \frac{1}{x} \right)}{x} = \lim_{x \to 0} \sin \left( \frac{1}{x} \right)
\]

The function \( \sin \left( \frac{1}{x} \right) \) oscillates between -1 and 1 as \( x \to 0 \), meaning the limit does not settle to a single value. Thus, the limit does not exist, and \( f'(0) \) does not exist for this function.

### Problem 60
The function is given by:

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

To find \( f'(0) \), we compute:

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

Now, \( x \sin \left( \frac{1}{x} \right) \) is bounded because \( \sin \left( \frac{1}{x} \right) \) oscillates between -1 and 1, so:

\[
- |x| \leq x \sin \left( \frac{1}{x} \right) \leq |x|
\]

As \( x \to 0 \), both \( -|x| \) and \( |x| \) approach 0. By the Squeeze Theorem:

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

Therefore, \( f'(0) = 0 \) for this function, so \( f'(0) \) exists and is equal to 0.

### Summary
1. For problem 59, \( f'(0) \) does not exist.
2. For problem 60, \( f'(0) = 0 \).

Would you like more details or have any questions on these calculations?

Here are 5 related questions to consider:
1. How does oscillation affect the existence of a limit?
2. What other conditions can make a limit exist or not?
3. Can you determine the continuity of these functions at \( x = 0 \)?
4. How would \( x^n \sin \left( \frac{1}{x} \right) \) behave for different values of \( n \)?
5. What if we replace \( \sin \left( \frac{1}{x} \right) \) with \( \cos \left( \frac{1}{x} \right) \) — would \( f'(0) \) still exist?

**Tip:** The Squeeze Theorem is particularly useful for analyzing limits of functions that oscillate but are bounded by functions approaching zero.

Determine whether f'(0) exists for the following functions:

59. f(x) = {x * sin(1/x) if x ≠ 0; 0 if x = 0}

60. f(x) = {x^2 * sin(1/x) if x ≠ 0; 0 if x = 0}

Explore whether f'(0) exists for two functions involving oscillating terms x * sin(1/x) and x^2 * sin(1/x). The solution uses limit definitions, differentiability concepts, and the Squeeze Theorem to analyze the functions at x = 0. Suitable for students in undergraduate calculus courses.

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