Question

Let \( f(x, y) = \begin{cases} \dfrac{xy}{(x^2 + y^2)^{\alpha}}, & (x, y) \neq (0,0) \\ 0, & (x, y) = (0,0) \end{cases} \) Then which one of the following is TRUE for \( f \) at the point \( (0, 0) \)?
 

• A. For \( \alpha = 1\), \( f \) is continuous but not differentiable
• B. For \( \alpha = \frac{1}{2} \), \( f \) is continuous and differentiable
• C. For \( \alpha = \frac{1}{4} \), \( f \) is continuous and differentiable
• D. For \( \alpha = \frac{3}{4} \), \( f \) is neither continuous nor differentiable

Hint:

For functions involving powers of \( x^2 + y^2 \), the value of \( \alpha \) is critical in determining the continuity and differentiability of the function at the origin.

Answer 1

The Correct Option is C

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

For continuity, we need: $$\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0) = 0$$

For $(x,y) \neq (0,0)$: $$\left|f(x,y)\right| = \left|\frac{xy}{(x^2+y^2)^\alpha}\right|$$

Using the inequality $|xy| \leq \frac{x^2+y^2}{2}$: $$\left|f(x,y)\right| \leq \frac{x^2+y^2}{2(x^2+y^2)^\alpha} = \frac{1}{2}(x^2+y^2)^{1-\alpha}$$

For the limit to be 0 as $(x,y) \to (0,0)$, we need: $$1 - \alpha > 0 \implies \alpha < 1$$

Continuity condition: $\alpha < 1$

Step 2: Check Differentiability at $(0,0)$

If $f$ is differentiable at $(0,0)$, then the partial derivatives must exist and satisfy: $$\lim_{(h,k) \to (0,0)} \frac{f(h,k) - f(0,0) - f_x(0,0)h - f_y(0,0)k}{\sqrt{h^2+k^2}} = 0$$

First, find the partial derivatives at $(0,0)$: $$f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$$ $$f_y(0,0) = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} = \lim_{k \to 0} \frac{0 - 0}{k} = 0$$

For differentiability, we need: $$\lim_{(h,k) \to (0,0)} \frac{f(h,k)}{\sqrt{h^2+k^2}} = 0$$

$$\left|\frac{f(h,k)}{\sqrt{h^2+k^2}}\right| = \left|\frac{hk}{(h^2+k^2)^\alpha \cdot \sqrt{h^2+k^2}}\right| = \frac{|hk|}{(h^2+k^2)^{\alpha + 1/2}}$$

Using $|hk| \leq \frac{h^2+k^2}{2}$: $$\left|\frac{f(h,k)}{\sqrt{h^2+k^2}}\right| \leq \frac{1}{2}(h^2+k^2)^{1-\alpha-1/2} = \frac{1}{2}(h^2+k^2)^{1/2-\alpha}$$

For this to approach 0, we need: $$\frac{1}{2} - \alpha > 0 \implies \alpha < \frac{1}{2}$$

Differentiability condition: $\alpha < \frac{1}{2}$

Step 3: Analyze Each Option

(A) $\alpha = 1$:

• Continuity: Need $\alpha < 1$ ✗ (NOT continuous)
• This option claims continuous but not differentiable - FALSE

(B) $\alpha = \frac{1}{2}$:

• Continuity: Need $\alpha < 1$ ✓ (continuous)
• Differentiability: Need $\alpha < \frac{1}{2}$ ✗ (NOT differentiable)
• This option claims continuous and differentiable - FALSE

(C) $\alpha = \frac{1}{4}$:

• Continuity: $\frac{1}{4} < 1$ ✓ (continuous)
• Differentiability: $\frac{1}{4} < \frac{1}{2}$ ✓ (differentiable)
• This option claims continuous and differentiable - TRUE 

(D) $\alpha = \frac{3}{4}$:

• Continuity: $\frac{3}{4} < 1$ ✓ (continuous)
• Differentiability: $\frac{3}{4} < \frac{1}{2}$ ✗ (NOT differentiable)
• This option claims neither continuous nor differentiable - FALSE

Answer: (C)