# Show a function is not continuous at a point

$$f(x,y) = \begin{cases} \dfrac{x^2 y^4}{x^4 + 6y^8}, & \text{if }(x,y)\neq(0,0) \\ 0, & \text{if }(x,y)=(0,0) \end{cases}$$

For the definition of differentiability, I have:

$$\lim_{h \rightarrow 0} \dfrac{||f(x+h, y+h) - f(x,y) - J(h)||}{||h||} = 0$$

So plugging in the function at $(0,0)$:

$$\lim_{h \rightarrow 0} \dfrac{\left\|\frac{(0+h)^2 (0+h)^4}{(0+h)^4 + 6(0+h)^8} - f(0,0) - J(h)\right\|}{||h||} = 0$$

Because $f(0,0)$ is defined to be 0 at that point:

$$\lim_{h \rightarrow 0} \dfrac{\left\|\frac{h^6}{h^4 + 6h^8} - J(h)\right\|}{||h||} = 0$$

I know $J(h)$ will be a $1 \times 2$ matrix, so I'm not sure how to deal with this situation here on out.

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Are you trying to show that it is not continuous at 0, or are you trying to show that it is not differentiable at 0? –  Brad Mar 23 at 17:54
Well, I am trying to show it is not continuous, so that consequently is not differentiable. The question specifically states to prove it is not continuous first. –  nabla blah Mar 23 at 17:55
There are functions which are continuous but not differentiable. For example, the absolute value function is continuous but not differentiable at 0. To show that a function is not continuous, then, it is not enough to show that it is not differentiable. –  Brad Mar 23 at 17:56
The exact wording is: "show that $f(0,0)$ is not differentiable by showing $f(0,0)$ is not continuous." I thought the work I was doing would be showing that the function is not continuous, but upon consideration, I guess I'm nowhere near close to the solution. I will have to start over. –  nabla blah Mar 23 at 17:57
The problem then is to show that $f$ is not continuous. The point is that this fact then implies that $f$ is not differentiable, because every differentiable function is continuous. –  Brad Mar 23 at 18:00

Hint: (i) Approach $(0,0)$ along the curve $x=y^2$; (ii) Approach $(0,0)$ along the curve $x=y$.
Is it also valid to approach $(0,0)$ along the curve $x = 0$ or $y = 0$? Or is that not allowed because we are trying to approach $(0,0)$? –  nabla blah Mar 23 at 18:06
Yes, for (ii) you could approach along $x=0$ or $y=0$. The limits for each choice are $0$, just as with my choice $x=y$. We do need something like (i) to conclude that the limit does not exist. –  André Nicolas Mar 23 at 18:10