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

  • $\begingroup$ You should type the problem into your post, not link to an image. $\endgroup$ – Mark Fantini Sep 13 '15 at 9:04
  • $\begingroup$ Yeah sorry i'm quite poor with latex and the problem is a Piecewise function making it rather hard to type up. $\endgroup$ – James Sep 13 '15 at 9:13
  • $\begingroup$ Please, explain how you get II) and III). And review the formal definition of partial derivative. $\endgroup$ – Miguel Sep 13 '15 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.

For the fourth, use the definition of partial derivatives:

$$\frac{\partial f}{\partial x}\bigg\vert_{(0,0)} := \lim_{h \to 0} \frac{f(0+h,0) - f(0,0)}{h},$$

$$\frac{\partial f}{\partial y}\bigg\vert_{(0,0)} := \lim_{h \to 0} \frac{f(0,0+h) - f(0,0)}{h}.$$

In light of this, think about the fifth question.


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