# proving that the function is not continuous at the origin

Show that the following function has partial derivatives at the origin, but it is not even continuous there:

$$f(x, y) = \begin{cases} 1 & \text{if x = 0 or y = 0} \\ 0 & \text{otherwise} \end{cases}$$

For the first part I took limits to find partial derivatives: $$\frac{\partial f(x,y)}{\partial x} = \lim_{h\to 0} \frac{f(0,h)-f(0,0)} h = 0$$ $$\frac{\partial f(x,y)}{\partial y} = \lim_{h\to 0} \frac{f(h,0)-f(0,0)} h = 0$$ and I got stuck with the second part.

• Welcome to the pit of despair. We like to see the despair in detail before we give out those free ladders you may have heard of. Please tell us what you have tried and where you got stuck. That way we will be able to target out answers to the appropriate level and specific difficulty. – dfeuer Nov 11 '13 at 2:39
• for the first part i took limits to find partial derivatives: df/dx= limit f(0,h)-f(0,0)/h= 0 h--->0 df/dy = limit f(h,0)-f(0,0)/h = 0 h--> 0 and i got stuck with the second part – user107929 Nov 11 '13 at 2:43

We can show the limit does not exist at $(0, 0)$ by going to $(0, 0)$ along two different paths. Along $y = 0$, we have $$\lim_{x \to 0} f(x, 0) = \lim_{x \to 0} 1 = 1.$$ But going along $y = x$, we have $$\lim_{x \to 0} f(x, x) = \lim_{x \to 0} 0 = 0.$$ So the limit $$\lim_{(x, y) \to (0, 0)} f(x, y)$$ does not exist and so $f$ is not continuous at $(0, 0)$.
Let $\epsilon = \frac 1 3$ and let $\delta>0$.
• @user107929 No, if both $x$ and $y$ are $0$, then its certainly true that one of $x$ and $y$ is $0$. So $f(x, y) = 1$. – Pratyush Sarkar Nov 11 '13 at 3:27
• @user107929 Because in any ball around $(0, 0)$, there are always points where the function is $0$ and also points where the function is $1$. So in fact the limit of the function at $(0, 0)$ does not even exist. – Pratyush Sarkar Nov 11 '13 at 3:31