To show that the limit $\lim_{(x,y) \to (0,0)} \frac{2xy^2}{x^2 + y^4}$ does not exist, we need to demonstrate that the value of the limit depends on the path taken as $(x, y)$ approaches $(0,0)$.

1. Path 1: $y = 0$

If we approach along the $x$-axis ($y = 0$): $\frac{2xy^2}{x^2 + y^4} = \frac{2x \cdot 0^2}{x^2 + 0^4} = 0.$ So, the limit along this path is $0$.

2. Path 2: $x = 0$

If we approach along the $y$-axis ($x = 0$): $\frac{2xy^2}{x^2 + y^4} = \frac{2 \cdot 0 \cdot y^2}{0 + y^4} = 0.$ So, the limit along this path is also $0$.

3. Path 3: $y = mx$ (where $m$ is a constant)

Now, let's approach along a line $y = mx$ (where $m$ is a constant): $\frac{2x(mx)^2}{x^2 + (mx)^4} = \frac{2x \cdot m^2 x^2}{x^2 + m^4 x^4} = \frac{2m^2 x^3}{x^2(1 + m^4 x^2)} = \frac{2m^2 x}{1 + m^4 x^2}.$ As $(x, y) \to (0,0)$, $x \to 0$, this expression approaches $0$.

4. Path 4: $y = x^{1/2}$

Now, let's try approaching along $y = x^{1/2}$: $\frac{2x(x^{1/2})^2}{x^2 + (x^{1/2})^4} = \frac{2x \cdot x}{x^2 + x^2} = \frac{2x^2}{2x^2} = 1.$ So, along this path, the limit is $1$.

Conclusion

Since the limit is $0$ along some paths and $1$ along another, the value of the limit depends on the path taken to approach $(0,0)$. Therefore, the limit does not exist.

Would you like further details on this topic, or have any questions?

Here are 5 related questions to explore further:

1. What is the definition of a limit in multiple dimensions?
2. How do we determine if a limit exists in higher dimensions?
3. What are common methods to show that a limit does not exist?
4. Can a limit exist if the function is not defined at the point of interest?
5. How do polar coordinates help in evaluating multivariable limits?

Tip: When evaluating limits in two variables, it is useful to try approaching the point along different paths to see if the limit is consistent. If not, the limit does not exist.