# Multivariable calc limits and continuity

Let f(x,y) = \left\{ \begin{aligned} 0, & \quad xy \neq 0, \\ 1, & \quad xy = 0. \end{aligned} \right.

1. Describe the surface $z=f(x,y)$.
2. Find the limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$ along the line $y=x$.
3. Is $f$ continuous at $(0,0)$?
4. Find $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$ at $(0,0)$, if they exist.
5. In single variable calculus, differentiability implies continuity. What conclusion, if any, can you make about the existence of partial derivatives and continuity for a function $f(x,y)$?

Could I get some help with question 2 to 5? Especially with the 4th question. I feel like i'm getting the answers but without knowing why, any help much appreciated.

my answers II) lim(0,0) along y=x = o

III)function is not continuous

IIII) not sure for this question

IIIII) Differentiability of a function does not necessarily imply that the function is continuous i'm guessing

• You should type the problem into your post, not link to an image. Sep 13, 2015 at 9:04
• Yeah sorry i'm quite poor with latex and the problem is a Piecewise function making it rather hard to type up. Sep 13, 2015 at 9:13
• Please, explain how you get II) and III). And review the formal definition of partial derivative. Sep 13, 2015 at 9:31

You should prove that $f$ is not continuous at the origin by finding another path with a different limit. Use the coordinate axis.
$$\frac{\partial f}{\partial x}\bigg\vert_{(0,0)} := \lim_{h \to 0} \frac{f(0+h,0) - f(0,0)}{h},$$
$$\frac{\partial f}{\partial y}\bigg\vert_{(0,0)} := \lim_{h \to 0} \frac{f(0,0+h) - f(0,0)}{h}.$$