We are tasked with evaluating the limit of the function:

$f(x, y) = \frac{xy + y^3}{x^2 + y^2}$

as $(x, y) \to (0, 0)$, along different paths.

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(a) Evaluate the limit along the x-axis, $y = 0$.

Substitute $y = 0$ into the function:

$f(x, 0) = \frac{x(0) + 0^3}{x^2 + 0^2} = \frac{0}{x^2} = 0.$

Thus, the limit along the x-axis is:

$\lim_{x \to 0} f(x, 0) = 0.$

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(b) Evaluate the limit along the y-axis, $x = 0$.

Substitute $x = 0$ into the function:

$f(0, y) = \frac{0(y) + y^3}{0^2 + y^2} = \frac{y^3}{y^2} = y.$

Thus, the limit along the y-axis is:

$\lim_{y \to 0} f(0, y) = 0.$

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(c) Evaluate the limit along the path $y = 2x$.

Substitute $y = 2x$ into the function:

$f(x, 2x) = \frac{x(2x) + (2x)^3}{x^2 + (2x)^2} = \frac{2x^2 + 8x^3}{x^2 + 4x^2} = \frac{2x^2 + 8x^3}{5x^2}.$

Simplifying this expression:

$f(x, 2x) = \frac{2x^2(1 + 4x)}{5x^2} = \frac{2(1 + 4x)}{5}.$

Taking the limit as $x \to 0$:

$\lim_{x \to 0} f(x, 2x) = \frac{2(1 + 0)}{5} = \frac{2}{5}.$

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Final Evaluation

The limits along different paths yield different results:

• Along the x-axis: $0$.
• Along the y-axis: $0$.
• Along the path $y = 2x$: $\frac{2}{5}$.

Since the limit depends on the path, the overall limit does not exist (DNE).

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Would you like further details on any part? Here are some related questions to consider:

1. What if we chose a different path, such as $y = x^2$?
2. How does path dependency affect the existence of multivariable limits?
3. Could polar coordinates help analyze this limit further?
4. Can we use L'Hopital's Rule in this context?
5. How would we graphically interpret this limit?

Tip: Multivariable limits may depend on the path, so testing along multiple paths is crucial when determining if a limit exists.