The problem provided is:

$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^3 + 2y - 3x}$

Solution:

To solve this limit as $(x, y) \to (0, 0)$, we'll check if the limit exists by trying different paths. If the limit yields different results along different paths, then the limit does not exist.

Path 1: Along $y = 0$

When $y = 0$, the expression becomes:

$\frac{x(0)^2}{x^3 + 2(0) - 3x} = \frac{0}{x^3 - 3x} = 0$

Thus, along $y = 0$, the limit is $0$.

Path 2: Along $x = 0$

When $x = 0$, the expression becomes:

$\frac{0 \cdot y^2}{0^3 + 2y - 3(0)} = \frac{0}{2y} = 0$

Thus, along $x = 0$, the limit is also $0$.

Path 3: Along $y = x$

Substitute $y = x$ into the expression:

$\frac{x(x)^2}{x^3 + 2x - 3x} = \frac{x^3}{x^3 - x} = \frac{x^3}{x(x^2 - 1)} = \frac{x^2}{x^2 - 1}$

Now, as $x \to 0$:

$\frac{x^2}{x^2 - 1} = \frac{0}{-1} = 0$

So, along $y = x$, the limit is also $0$.

Conclusion:

Since the limit along the different paths gives the same result of $0$, we can conclude that:

$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^3 + 2y - 3x} = 0$

Would you like further details or explanations on any part of the solution?

Related Questions:

1. How can polar coordinates be used to solve multivariable limits?
2. What happens if we take the limit along more complex paths like $y = x^2$?
3. How do you determine when a limit does not exist in multivariable calculus?
4. What are the common techniques to evaluate limits in two variables?
5. Can L'Hôpital's rule be applied to multivariable limits, and if so, how?

Tip:

When testing for limits in two variables, always try multiple paths. If the limit differs depending on the path, the limit does not exist.