The limit problem in the image is as follows:

$\lim_{(x, y) \to (0,0)} \frac{x \cdot y^4}{x^4 + y^4}$

To evaluate this limit, we need to check if it tends to the same value along different paths as $(x, y)$ approaches $(0, 0)$. If the limit depends on the path, the overall limit does not exist.

Step 1: Evaluate the limit along $y = 0$

If $y = 0$, the expression becomes: $\frac{x \cdot 0^4}{x^4 + 0^4} = 0$ So, along the path $y = 0$, the limit is 0.

Step 2: Evaluate the limit along $x = 0$

If $x = 0$, the expression becomes: $\frac{0 \cdot y^4}{0^4 + y^4} = 0$ So, along the path $x = 0$, the limit is also 0.

Step 3: Evaluate the limit along $y = x$

If we substitute $y = x$, the expression becomes: $\frac{x \cdot x^4}{x^4 + x^4} = \frac{x^5}{2x^4} = \frac{x}{2}$ As $x \to 0$, $\frac{x}{2} \to 0$.

Step 4: Evaluate the limit along $y = -x$

If we substitute $y = -x$, the expression becomes: $\frac{x \cdot (-x)^4}{x^4 + (-x)^4} = \frac{x \cdot x^4}{x^4 + x^4} = \frac{x^5}{2x^4} = \frac{x}{2}$ Again, as $x \to 0$, $\frac{x}{2} \to 0$.

Conclusion:

Since the limit along multiple paths (e.g., $y = 0$, $x = 0$, $y = x$, and $y = -x$) all lead to the same result of 0, we conclude that the limit exists and is equal to 0.

Thus, the limit is: $\boxed{0}$

Would you like more details on any of the steps?

Here are 5 related questions:

1. How do you evaluate limits in multivariable calculus using polar coordinates?
2. What are the common strategies to prove a limit does not exist?
3. How does the limit change when evaluated along different types of curves like $y = x^2$?
4. How would this limit behave if the exponents in the denominator were different?
5. What are some common pitfalls when evaluating limits in two variables?

Tip: When solving limits involving two variables, always check the behavior along multiple paths to ensure the limit is consistent.