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The image shows two limit problems. Let's solve them step by step.

### (a) \(\lim_{x \to 0} \frac{x - \sin(x)}{x^3}\)

#### Step 1: Apply L'Hopital's Rule
This limit is an indeterminate form \(\frac{0}{0}\) as both the numerator \(x - \sin(x)\) and the denominator \(x^3\) approach 0 as \(x \to 0\). We can apply L'Hopital's Rule, which states that if the limit is of the form \(\frac{0}{0}\), we can take the derivatives of the numerator and the denominator.

- Derivative of the numerator: \(\frac{d}{dx}(x - \sin(x)) = 1 - \cos(x)\)
- Derivative of the denominator: \(\frac{d}{dx}(x^3) = 3x^2\)

Thus, the limit becomes:
\[
\lim_{x \to 0} \frac{1 - \cos(x)}{3x^2}
\]

#### Step 2: Apply L'Hopital's Rule Again
This is still of the form \(\frac{0}{0}\), so we apply L'Hopital's Rule again:

- Derivative of the numerator: \(\frac{d}{dx}(1 - \cos(x)) = \sin(x)\)
- Derivative of the denominator: \(\frac{d}{dx}(3x^2) = 6x\)

Now the limit becomes:
\[
\lim_{x \to 0} \frac{\sin(x)}{6x}
\]

#### Step 3: Solve the Limit
We know that \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\), so the expression simplifies to:
\[
\frac{1}{6}
\]

Thus, the solution to (a) is:
\[
\boxed{\frac{1}{6}}
\]

---

### (b) \(\lim_{x \to \infty} \frac{x^3 \cos(x^2)}{3x^4 + 2}\)

#### Step 1: Analyze the Degree of Terms
As \(x \to \infty\), the denominator grows as \(3x^4\), which dominates over the constant term 2. In the numerator, we have \(x^3 \cos(x^2)\). The cosine function oscillates between -1 and 1, meaning the numerator's magnitude is bounded by \(-x^3\) and \(x^3\).

#### Step 2: Estimate the Limit
We can think of \(\cos(x^2)\) as bounded while \(x^3\) grows large. So, the numerator behaves like \(x^3\), and the denominator behaves like \(3x^4\). This suggests that:

\[
\lim_{x \to \infty} \frac{x^3 \cos(x^2)}{3x^4 + 2} \approx \lim_{x \to \infty} \frac{x^3}{3x^4} = \lim_{x \to \infty} \frac{1}{3x} = 0
\]

Thus, the solution to (b) is:
\[
\boxed{0}
\]

---

Would you like further details or clarifications? Here are five additional related questions for further exploration:

1. How is L'Hopital's Rule used in cases involving \(\frac{0}{0}\) indeterminate forms?
2. Can L'Hopital's Rule be applied when the limit is of the form \(\frac{\infty}{\infty}\)?
3. How would you evaluate a limit if the denominator grows much faster than the numerator?
4. What are other techniques for solving limits involving trigonometric functions besides L'Hopital's Rule?
5. How do the bounds on \(\cos(x)\) help simplify limits when it oscillates?

**Tip:** When applying L'Hopital's Rule, make sure the limit is in an indeterminate form before differentiating the numerator and denominator.

Find the following limits: (a) lim(x -> 0) [(x - sin(x)) / x^3], (b) lim(x -> infinity) [(x^3 cos(x^2)) / (3x^4 + 2)]

Learn how to solve limits involving trigonometric functions and apply L'Hopital's Rule. This problem set includes finding limits such as lim(x -> 0) [(x - sin(x)) / x^3] and lim(x -> infinity) [(x^3 cos(x^2)) / (3x^4 + 2)]. Suitable for advanced high school or college-level calculus students.

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