The given expression is:

$\lim_{(x, y) \to (0, 0)} \frac{y^5}{x^4 + y^2}$

To determine whether this limit exists, we need to check if the limit value depends on the path taken towards the origin $(0, 0)$.

Approach 1: Along the $x$-axis ($y = 0$)

Substitute $y = 0$ into the expression:

$\frac{0^5}{x^4 + 0^2} = \frac{0}{x^4} = 0$

So, along the $x$-axis, the limit is 0.

Approach 2: Along the $y$-axis ($x = 0$)

Substitute $x = 0$ into the expression:

$\frac{y^5}{0^4 + y^2} = \frac{y^5}{y^2} = y^3$

As $(x, y) \to (0, 0)$, $y^3 \to 0$. So, along the $y$-axis, the limit is also 0.

Approach 3: Along the line $y = mx$ (for some constant $m$)

Substitute $y = mx$ into the expression:

$\frac{(mx)^5}{x^4 + (mx)^2} = \frac{m^5 x^5}{x^4 + m^2 x^2} = \frac{m^5 x^5}{x^2(x^2 + m^2)}$

Factor out $x^2$:

$= \frac{m^5 x^3}{x^2 + m^2}$

As $x \to 0$, the numerator approaches $0$, and the denominator approaches $m^2$, so the limit is:

$\lim_{x \to 0} \frac{m^5 x^3}{x^2 + m^2} = 0$

Conclusion:

Based on these paths (the $x$-axis, the $y$-axis, and the line $y = mx$), the limit appears to be $0$ along each approach. Hence, we conclude:

$\lim_{(x, y) \to (0, 0)} \frac{y^5}{x^4 + y^2} = 0$

If you'd like more details or another approach to this problem, feel free to ask!

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Related Questions:

1. How do you check the limit of a multivariable function using polar coordinates?
2. What happens if the numerator or denominator has a higher degree than the other in such limits?
3. What does it mean if the limit depends on the path taken?
4. How can you apply the epsilon-delta definition of limits in multivariable functions?
5. How does the squeeze theorem apply to limits involving two variables?

Tip:

When checking multivariable limits, testing along various paths is a good initial approach. If the limit value changes based on the path, the limit does not exist.