Question

Let $f ∶ R^2 → R$ be a function defined as
$f(x,y) =\begin{cases}\frac{x^2y}{x^4 + Y^2} & if (x, y) \neq (0, 0)\\ 0 & if (x, y) = (0, 0)\end{cases} $ Then, which of the following is/are CORRECT?

• A. $ \lim_{(x,y) \rightarrow (0,0)} 𝑓(x, y) = 0$
• B. $f_x(0, 0) = 0$
• C. 𝑓(x, y) is not continuous at (0, 0)
• D. Both $f_x$ and $f_y$ do not exist at (0, 0)

Answer 1

The Correct Option is B, C

Analysis of Function f(x, y) at (0,0)

Given Function: 

The function f(x, y) is defined as:

f(x, y) = (x²y) / (x⁴ + y²), if (x, y) ≠ (0, 0)

f(x, y) = 0, if (x, y) = (0, 0)

Step 1: Checking Continuity at (0, 0)

To check continuity, we evaluate the limit lim(x,y)→(0,0) f(x, y) along different paths:

• Along y = 0:

f(x, 0) = (x² × 0) / (x⁴ + 0²) = 0.

• Along x = 0:

f(0, y) = (0² × y) / (0⁴ + y²) = 0.

• Along y = mx²:

f(x, mx²) = (x² × mx²) / (x⁴ + (mx²)²) = (m x⁴) / (x⁴ + m² x⁴) = m / (1 + m²).

Since the limit depends on m, it varies along different paths, meaning lim(x,y)→(0,0) f(x, y) does not exist.

Conclusion: The function is not continuous at (0, 0), so Option (C) is correct.

Step 2: Partial Derivative with Respect to x at (0, 0)

The partial derivative fₓ(0, 0) is defined as:

fₓ(0, 0) = lim(h→0) (f(h, 0) − f(0, 0)) / h

Since f(h, 0) = 0 and f(0, 0) = 0, we get:

fₓ(0, 0) = lim(h→0) (0 - 0) / h = 0.

Conclusion: Option (B) is correct.

Step 3: Partial Derivative with Respect to y at (0, 0)

The partial derivative fᵧ(0, 0) is defined as:

fᵧ(0, 0) = lim(h→0) (f(0, h) − f(0, 0)) / h

Since f(0, h) = 0 and f(0, 0) = 0, we get:

fᵧ(0, 0) = lim(h→0) (0 - 0) / h = 0.

Since both fₓ(0, 0) and fᵧ(0, 0) exist and are equal to 0, Option (D) is incorrect.

Step 4: Evaluating the Limit of f(x, y) at (0, 0)

Since the limit of f(x, y) depends on the path taken (as shown in Step 1), lim(x,y)→(0,0) f(x, y) does not exist.

Conclusion: Option (A) is incorrect.

Final Answer:

The correct options are:

• (B) Partial derivative fₓ(0, 0) = 0
• (C) The function is not continuous at (0, 0)