limits

Question

Let $f(x,y) = \begin{cases} 1, & \quad if \: xy=0, \\ xy, & \quad xy \neq 0. \end{cases}$

Then

• A. $f$ is continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ exists
• B. $f$ is not continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ exists
• C. $f$ is continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ does not exist
• D. $f$ is not continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ does not exist

Answer 1

D is the ans.
proof for continuity at 0,0
Assuming x = rcos⊖ and y= rsin⊖ (Polar coordinates)
f(x,y) =  $lim x,y\rightarrow (0,0)$ xy
= $limr\rightarrow 0$  rcos⊖ * rsin⊖
= 0
F(0,0) = 1
Since RHL ≄ F(0,0)
thus not continuous at 0,0.