Which one of the following statements is TRUE:
f left parenthesis x comma y right parenthesis equals fraction numerator x squared y over denominator x to the power of 4 plus y squared end fraction
a.
along the line y equals negative x comma   limit as left parenthesis x comma y right parenthesis rightwards arrow left parenthesis 0 comma 0 right parenthesis of   f left parenthesis x comma y right parenthesis equals negative 1
b.
limit as left parenthesis x comma y right parenthesis rightwards arrow left parenthesis 0 comma 0 right parenthesis of   f left parenthesis x comma y right parenthesis equals 0
c.
limit as left parenthesis x comma y right parenthesis rightwards arrow left parenthesis 0 comma 0 right parenthesis of   f left parenthesis x comma y right parenthesis equals 1 half
d.
along the parabola y equals x squared comma   limit as left parenthesis x comma y right parenthesis rightwards arrow left parenthesis 0 comma 0 right parenthesis of   f left parenthesis x comma y right parenthesis equals 2
e.
along the line y equals 2 x comma   limit as left parenthesis x comma y right parenthesis rightwards arrow left parenthesis 0 comma 0 right parenthesis of   f left parenthesis x comma y right parenthesis equals 0

To determine which statement is true, we need to analyze the limit of the function 

\[
f(x, y) = \frac{x^2 y}{x^4 + y^2}
\]

as \((x, y)\) approaches \((0, 0)\) along various paths.

### A) Along the line \(y = -x\):
Substituting \(y = -x\):

\[
f(x, -x) = \frac{x^2(-x)}{x^4 + (-x)^2} = \frac{-x^3}{x^4 + x^2} = \frac{-x^3}{x^2(x^2 + 1)} = \frac{-x}{x^2 + 1}
\]

As \(x \to 0\):

\[
\lim_{x \to 0} \frac{-x}{x^2 + 1} = 0
\]

### B) Limit as \((x, y) \to (0, 0)\):
We can check various paths to see if we approach different limits.

### C) Along the line \(y = 0\):
\[
f(x, 0) = \frac{x^2(0)}{x^4 + 0} = 0
\]

### D) Along the line \(x = 0\):
\[
f(0, y) = \frac{0}{0 + y^2} = 0
\]

### E) Along the parabola \(y = x^2\):
Substituting \(y = x^2\):

\[
f(x, x^2) = \frac{x^2(x^2)}{x^4 + (x^2)^2} = \frac{x^4}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2}
\]

As \(x \to 0\):

\[
\lim_{x \to 0} f(x, x^2) = \frac{1}{2}
\]

### F) Along the line \(y = 2x\):
Substituting \(y = 2x\):

\[
f(x, 2x) = \frac{x^2(2x)}{x^4 + (2x)^2} = \frac{2x^3}{x^4 + 4x^2} = \frac{2x^3}{x^2(x^2 + 4)} = \frac{2x}{x^2 + 4}
\]

As \(x \to 0\):

\[
\lim_{x \to 0} \frac{2x}{x^2 + 4} = 0
\]

### Summary of Limits:
- **A)** Limit along \(y = -x\) is \(0\).
- **B)** Overall limit as \((x,y) \to (0,0)\) has multiple paths yielding different values.
- **C)** Limit along \(y = 0\) is \(0\).
- **D)** Limit along the parabola \(y = x^2\) is \(\frac{1}{2}\).
- **E)** Limit along \(y = 2x\) is \(0\).

The true statement is:

- **C)**: \( \lim_{(x,y) \to (0,0)} f(x,y) = 0\) is the limit along \(y = 0\).

However, since the limit depends on the path and does not converge to a single value, the most consistent conclusion is that **D)** along the parabola \(y = x^2\) gives \(\frac{1}{2}\).

Therefore, the answer is **C)** or **D)**, based on paths that yield consistent results.

Do you want more details or have any questions? Here are some follow-up questions you might consider:
1. What are other paths to explore for limits?
2. How can we show that the limit does not exist?
3. What is the significance of limits in multivariable calculus?
4. How does this function behave for larger values of \(x\) and \(y\)?
5. Can we apply the epsilon-delta definition of a limit here? 

**Tip:** When exploring limits in multivariable functions, it's important to test multiple paths to confirm whether a limit exists.

Which one of the following statements is TRUE: f(x, y) = x^2y / (x^4 + y^2) ...

Explore the limits of the multivariable function f(x, y) = x^2y / (x^4 + y^2) as (x, y) approaches (0, 0). Detailed solutions for various paths are provided, helping to determine which statements about the limit are true. Suitable for advanced students in Grades 11-12.

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