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The image contains four limit problems. Let's work through them one by one.

### Problem 8:
\[
\lim_{(x, y) \to (0, 0)} \frac{1 - \cos(x^2 + y^2)}{2(x^2 + y^2)}
\]

#### Solution:
1. Let \( r^2 = x^2 + y^2 \). Then as \( (x, y) \to (0, 0) \), \( r \to 0 \).
2. The given expression simplifies to:
   \[
   \lim_{r \to 0} \frac{1 - \cos(r^2)}{2r^2}
   \]
3. Use the approximation \( 1 - \cos(z) \approx \frac{z^2}{2} \) for small \( z \):
   \[
   \frac{1 - \cos(r^2)}{2r^2} \approx \frac{r^4 / 2}{2r^2} = \frac{r^2}{4}
   \]
4. As \( r \to 0 \), \( \frac{r^2}{4} \to 0 \).
   
Thus, the limit is:
\[
\boxed{0}
\]

### Problem 9:
\[
\lim_{(x, y) \to (0, 0)} \frac{x^4 - 16y^4}{x^2 - 4y^2}
\]

#### Solution:
1. First, check if direct substitution \( (x, y) = (0, 0) \) results in an indeterminate form:
   \[
   \frac{0^4 - 16(0^4)}{0^2 - 4(0^2)} = \frac{0}{0}
   \]
   So, it's indeterminate, and we need another method.
   
2. Try approaching along different paths:
   - Path 1: Along \( y = 0 \):
     \[
     \frac{x^4 - 16(0)}{x^2 - 4(0)} = \frac{x^4}{x^2} = x^2
     \]
     As \( x \to 0 \), this goes to 0.
   - Path 2: Along \( x = 0 \):
     \[
     \frac{0^4 - 16y^4}{0^2 - 4y^2} = \frac{-16y^4}{-4y^2} = 4y^2
     \]
     As \( y \to 0 \), this also goes to 0.

Since both paths yield the same result, the limit is:
\[
\boxed{0}
\]

### Problem 10:
\[
\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}
\]

#### Solution:
1. Using polar coordinates, where \( r^2 = x^2 + y^2 \) and \( x = r \cos \theta \), \( y = r \sin \theta \):
   \[
   \frac{xy}{\sqrt{x^2 + y^2}} = \frac{r \cos \theta \cdot r \sin \theta}{r} = r \cos \theta \sin \theta
   \]
2. As \( r \to 0 \), the expression becomes \( r \cdot \cos \theta \sin \theta \), which tends to 0.

Thus, the limit is:
\[
\boxed{0}
\]

### Problem 11:
\[
\lim_{(x, y) \to (0, 0)} \frac{x^3 y^3}{x^2 + y^2}
\]

#### Solution:
1. Again, convert to polar coordinates:
   \[
   \frac{x^3 y^3}{x^2 + y^2} = \frac{r^3 \cos^3 \theta \sin^3 \theta}{r^2} = r \cos^3 \theta \sin^3 \theta
   \]
2. As \( r \to 0 \), this expression tends to 0 because \( r \cdot \cos^3 \theta \sin^3 \theta \to 0 \).

Thus, the limit is:
\[
\boxed{0}
\]

---

Would you like more details or have any further questions?

Here are some related questions to expand on this:
1. How do polar coordinates simplify evaluating two-variable limits?
2. What is L'Hopital's Rule, and when can it be applied in limit problems?
3. Can different approaches (paths) to a limit give different results? What does that mean?
4. How does the small-angle approximation help in limit problems involving trigonometric functions?
5. What are common indeterminate forms encountered in limit problems?

**Tip:** When working with limits approaching a point in two or more variables, checking along different paths is often a good starting point to find the behavior of the function.

Explore solutions to four challenging multivariable limit problems using polar coordinates, small-angle approximations, and limit laws. Topics covered include indeterminate forms and the application of trigonometric identities in solving limits. Suitable for college-level or advanced high school calculus students.

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