Question

Let the function \(f: R^2⇢R\)  be \(f(x, y) = \frac{xy^2}{ x^3+  2xy + y^3}\,\,  f(0, 0) = 0.\) Then

• A. f is differentiable at (0, 0).
• B. f, does not exist at (0, 0).
• C. does not exist at (0, 0).
• D. f is not continuous at (0, 0).

Answer 1

The Correct Option is D

To determine whether the function \(f: R^2 \to R\) defined as \(f(x, y) = \frac{xy^2}{x^3 + 2xy + y^3}\), with \(f(0, 0) = 0\), is continuous or differentiable at the point \((0, 0)\), we need to check the limit of the function as it approaches \((0, 0)\).

Step 1: Check Continuity at \((0, 0)\)

For the function to be continuous at \((0, 0)\), the limit as \((x, y) \to (0, 0)\) must equal \(f(0, 0)\), which is 0. Let's analyze the limit:

Consider approaching \((0, 0)\) along the line \(y = mx\):

\[\begin{align*} f(x, mx) &= \frac{x(mx)^2}{x^3 + 2x(mx) + (mx)^3} \\ &= \frac{mx^3}{x^3 + 2mx^2 + m^3x^3} \\ &= \frac{mx}{1 + 2m/x + m^3}. \end{align*}\]

As \(x \to 0\), this expression tends to \(\frac{0}{0}\), indicating we need a better approach. Let's explore further:

Rewrite the expression in polar coordinates \(x = r\cos\theta, y = r\sin\theta\), so:

\[f(r\cos\theta, r\sin\theta) = \frac{r\cos\theta (r\sin\theta)^2}{(r\cos\theta)^3 + 2(r\cos\theta)(r\sin\theta) + (r\sin\theta)^3}.\]

Simplifying, we get:

\[f(r\cos\theta, r\sin\theta) = \frac{r^3 \cos\theta \sin^2\theta}{r^3 (\cos^3\theta + 2\cos\theta\sin\theta + \sin^3\theta)}.\]

This simplifies to:

\[\frac{\cos\theta \sin^2\theta}{\cos^3\theta + 2\cos\theta\sin\theta + \sin^3\theta}.\]

The expression depends on \(\theta\), not \(r\). Hence, the limit is not consistently zero as \(r \to 0\) for every \(\theta\). Thus, \(f(x, y)\) is not continuous at \((0, 0)\).

Conclusion: The function \(f(x, y)\) is not continuous at the point \((0, 0)\). Therefore, the correct answer is: f is not continuous at (0, 0).