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The problem asks to find for which functions the limit \( \lim_{(x, y) \to (0,0)} f(x, y) \) does not exist. There are four functions provided:

a. \( f(x, y) = \frac{x^2 y^2}{x^4 + y^4} \)

b. \( f(x, y) = \frac{x^4 y^4}{(x^2 + y^4)^3} \)

c. \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \)

d. \( f(x, y) = \frac{x^2 + y}{x^2 + y^2} \)

### Approach:
For each function, we will check if the limit depends on the path taken towards the origin \((0, 0)\). If the limit is path-dependent, then the limit does not exist.

#### Function (a): \( f(x, y) = \frac{x^2 y^2}{x^4 + y^4} \)

- Along the path \( y = x \), we have:
  \[
  f(x, x) = \frac{x^2 x^2}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2}
  \]
  
- Along the path \( y = 0 \), we have:
  \[
  f(x, 0) = \frac{x^2 \cdot 0^2}{x^4 + 0^4} = 0
  \]
  
Since the limit depends on the path, the limit does **not** exist for function (a).

#### Function (b): \( f(x, y) = \frac{x^4 y^4}{(x^2 + y^4)^3} \)

- Along the path \( y = x \), we have:
  \[
  f(x, x) = \frac{x^4 x^4}{(x^2 + x^4)^3} = \frac{x^8}{(x^2 + x^4)^3}
  \]
  For small \( x \), \( x^4 \) becomes very small compared to \( x^2 \), so this tends to zero as \( x \to 0 \).

- Along the path \( y = 0 \), we have:
  \[
  f(x, 0) = \frac{x^4 \cdot 0^4}{(x^2 + 0^4)^3} = 0
  \]
  
Both paths give a limit of 0, so this limit **exists**.

#### Function (c): \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \)

- Along the path \( y = x \), we have:
  \[
  f(x, x) = \frac{x^2 - x^2}{x^2 + x^2} = 0
  \]
  
- Along the path \( y = 0 \), we have:
  \[
  f(x, 0) = \frac{x^2 - 0^2}{x^2 + 0^2} = 1
  \]
  
Since the limit depends on the path, the limit does **not** exist for function (c).

#### Function (d): \( f(x, y) = \frac{x^2 + y}{x^2 + y^2} \)

- Along the path \( y = x \), we have:
  \[
  f(x, x) = \frac{x^2 + x}{x^2 + x^2} = \frac{x^2 + x}{2x^2} = \frac{1}{2} + \frac{1}{2x}
  \]
  As \( x \to 0 \), this expression tends to infinity.

- Along the path \( y = 0 \), we have:
  \[
  f(x, 0) = \frac{x^2 + 0}{x^2 + 0^2} = 1
  \]

Since the limit depends on the path, the limit does **not** exist for function (d).

### Conclusion:
The limits do **not exist** for functions (a), (c), and (d).

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How do different paths affect the existence of a limit in multivariable functions?
2. Why does the path \( y = 0 \) often simplify calculations for limits in two variables?
3. Can you generalize the behavior of limits in two variables when there is symmetry in the function?
4. How would polar coordinates help in determining limits for these types of functions?
5. How do discontinuities impact the existence of limits?

**Tip:** When checking limits in multivariable functions, try multiple paths to identify potential path-dependence.

Find the functions where the limit \( \lim_{(x, y) \to (0,0)} f(x, y) \) does not exist for the given functions: a. \( f(x, y) = \frac{x^2 y^2}{x^4 + y^4} \), b. \( f(x, y) = \frac{x^4 y^4}{(x^2 + y^4)^3} \), c. \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \), d. \( f(x, y) = \frac{x^2 + y}{x^2 + y^2} \).

Explore how to determine the existence of multivariable limits for functions like \( f(x, y) = \frac{x^2 y^2}{x^4 + y^4} \), including concepts like path-dependence and using different approaches. Suitable for advanced calculus students.

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