Question

Let \(f: \mathbb{R}^2 \to \mathbb{R}\) be a function defined as follows:
\[ f(x,y) = \begin{cases} \dfrac{x\sqrt{x^2+y^2}}{|x|} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases} \]
Which of the following statements is/are TRUE?

• A. \(f(x,y)\) is continuous at \((0,0)\).
• B. For any \(u \in \mathbb{R}^2\), the directional derivative of \(f\) at \((0,0)\) in the direction of \(u\) exists.
• C. \(\dfrac{\partial f}{\partial x}(0,0) = 1 = \dfrac{\partial f}{\partial y}(0,0)\).
• D. \(f(x,y)\) is differentiable at \((0,0)\).

Hint:

Rewrite \(f(x,y) = \text{sign}(x)\sqrt{x^2+y^2}\) for \(x \neq 0\) and check whether this bound goes to zero along every path, and whether the directional derivative is linear in direction.

Answer 1

The Correct Option is A, B

A neat way to handle this problem is to switch to polar coordinates near the origin: write $x = r\cos\theta$, $y = r\sin\theta$ with $r \ge 0$. For $x \ne 0$ (that is, $\cos\theta \ne 0$),

$$f(r,\theta) = \frac{x\sqrt{x^2+y^2}}{|x|} = \frac{r\cos\theta \cdot r}{r|\cos\theta|} = r \cdot \text{sign}(\cos\theta)$$

and $f = 0$ when $\cos\theta = 0$ (the $y$-axis, where $x=0$ by definition). So near the origin, $f$ always looks like $r$ times a bounded factor of size at most $1$.

1. (A) Continuity at $(0,0)$: True. Since $|f(r,\theta)| \le r$ for every angle $\theta$, letting $r \to 0$ forces $f \to 0$, which matches $f(0,0)=0$. The angle plays no role in the limit, so the limit is the same from every direction.
2. (B) Directional derivative exists for every $u$: True. Along a ray at fixed angle $\theta$, $f(r,\theta) = r \cdot \text{sign}(\cos\theta)$ is linear in $r$, so its derivative in $r$ at $r=0$ is just the constant $\text{sign}(\cos\theta)$, a finite number for every angle (and $0$ when $\cos\theta = 0$).
3. (C) $\frac{\partial f}{\partial x}(0,0) = 1 = \frac{\partial f}{\partial y}(0,0)$: False. Taking $\theta = 0$ (the positive $x$-direction) gives $\text{sign}(\cos 0)=1$, so $\partial f/\partial x(0,0) = 1$. But taking $\theta = \pi/2$ (the $y$-direction) gives $\cos\theta = 0$, which falls in the $x=0$ branch where $f \equiv 0$, so $\partial f/\partial y(0,0) = 0$, not $1$.
4. (D) Differentiable at $(0,0)$: False. A differentiable function must have its directional derivative equal to $\cos\theta \cdot f_x(0,0) + \sin\theta \cdot f_y(0,0) = \cos\theta$, a smooth function of $\theta$. But the actual directional derivative here is $\text{sign}(\cos\theta)$, which jumps between $-1$ and $1$ and does not match $\cos\theta$ except at $\theta = 0, \pi$. This mismatch rules out differentiability.

So only (A) and (B) hold up.

Let's summarize:

• In polar form $f$ collapses to $r$ times a sign that depends only on the angle, which is enough to guarantee continuity and existence of every directional derivative.
• That same sign factor is not a smooth (linear) function of direction, which is exactly why the two partials are unequal and why differentiability fails.

The correct options are (A) and (B).

\[ \boxed{\text{(A) and (B)}} \]