Question

Let \(f(x, y) = \begin{cases} \frac{xy}{\sqrt{x^2 + y^2}}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases}\), then

• A. $\lim_{(x, y) \to (0, 0)} f(x, y)$ does not exist.
• B. $f(x, y)$ is differentiable at $(0, 0)$.
• C. $f(x, y)$ is continuous but not differentiable at $(0, 0)$.
• D. $\lim_{(x, y) \to (0, 0)} f(x, y)$ exists, but $f(x, y)$ is not continuous at $(0, 0)$.

Hint:

For functions of the form $f(x,y) = \frac{x^a y^b}{(x^2+y^2)^c}$:
- Continuous at $(0,0)$ if $a+b > 2c$.
- Differentiable at $(0,0)$ if $a+b > 2c + 1$.
Here $a=1, b=1, c=1/2$, so $a+b = 2 > 2(1/2) = 1$ (Continuous!), but $a+b = 2 \ngtr 1+1 = 2$ (Not Differentiable!).

Answer 1

The Correct Option is C

Step 1: Concept
For a multivariable function $f(x, y)$:
1. Continuity at $(0, 0)$: $\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0)$.
2. Partial Derivatives at $(0,0)$: \[ f_x(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h} \] \[ f_y(0, 0) = \lim_{k \to 0} \frac{f(0, k) - f(0, 0)}{k} \] 3. Differentiability at $(0, 0)$: $f$ is differentiable at $(0,0)$ if and only if: \[ \lim_{(h, k) \to (0, 0)} \frac{f(h, k) - f(0, 0) - h f_x(0, 0) - k f_y(0, 0)}{\sqrt{h^2 + k^2}} = 0 \]

Step 2: Key Formulas and Approach
We test continuity using polar coordinates ($x = r\cos\theta, y = r\sin\theta$), calculate partial derivatives $f_x(0,0)$ and $f_y(0,0)$, and then evaluate the differentiability limit.

Step 3: Step-by-step Explanation

• 1. Checking Continuity at $(0,0)$: Convert to polar coordinates: $x = r \cos\theta, y = r \sin\theta$. \[ f(r \cos\theta, r \sin\theta) = \frac{(r \cos\theta)(r \sin\theta)}{\sqrt{r^2 \cos^2\theta + r^2 \sin^2\theta}} = \frac{r^2 \cos\theta \sin\theta}{r} = r \cos\theta \sin\theta \] Now evaluate the limit as $r \to 0$: \[ \lim_{(x,y) \to (0,0)} f(x,y) = \lim_{r \to 0} (r \cos\theta \sin\theta) = 0 \] Since $\lim_{(x,y) \to (0,0)} f(x,y) = 0 = f(0,0)$, $f(x,y)$ is continuous at $(0, 0)$.

• 2. Computing Partial Derivatives at $(0,0)$: \[ f_x(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0 \] \[ f_y(0, 0) = \lim_{k \to 0} \frac{f(0, k) - f(0, 0)}{k} = \lim_{k \to 0} \frac{0 - 0}{k} = 0 \] So $f_x(0,0) = 0$ and $f_y(0,0) = 0$.

• 3. Testing Differentiability at $(0,0)$: Define the error term limit $L$: \[ L = \lim_{(h, k) \to (0, 0)} \frac{f(h, k) - f(0, 0) - h f_x(0, 0) - k f_y(0, 0)}{\sqrt{h^2 + k^2}} \] Substituting $f(0,0)=0, f_x(0,0)=0, f_y(0,0)=0$: \[ L = \lim_{(h, k) \to (0, 0)} \frac{\frac{hk}{\sqrt{h^2 + k^2}}}{\sqrt{h^2 + k^2}} = \lim_{(h, k) \to (0, 0)} \frac{hk}{h^2 + k^2} \] To evaluate this limit, approach along the line $k = mh$: \[ L = \lim_{h \to 0} \frac{h(mh)}{h^2 + (mh)^2} = \lim_{h \to 0} \frac{m h^2}{h^2(1 + m^2)} = \frac{m}{1 + m^2} \] The value of the limit depends on the slope $m$. For $m=1$, $L = 1/2$; for $m=0$, $L = 0$.
Since the limit depends on $m$, it is non-unique and therefore does not exist (and specifically is not zero).
Hence, $f(x, y)$ is not differentiable at $(0, 0)$.

Step 4: Final Answer
The function $f(x, y)$ is continuous at $(0, 0)$ but not differentiable at $(0, 0)$. Thus, Option (C) is correct.