Explain why continuity along straight lines is not enough to conclude continuity

Consider the function with domain $A = \{ (x,y) \in \, \mathbb{R}^2: (x,y) \neq (0,0)\}$ given by

$$\frac{2x^2y}{x^4+y^2}$$

Letting $(x,y)$ approach $(0,0)$ along the straight line $y=ax$ , where $a$ is a real constant, we find that the limit is zero. This is not enough to conclude that the limit exists. Explain why.

I'm incredbly confused so...

$$\lim_{x=y \to 0} \frac{2x^2y}{x^4+y^2} = \frac{0}{0} =0$$

Amongst two different paths...

$$\lim_{x \to 0} \frac{2x^2y}{x^4+y^2} = \frac{0\times y}{0+y^2} =0$$ $$\lim_{y \to 0} \frac{2x^2y}{x^4+y^2} = \frac{x\times 0}{x^4+0} =0$$ Which works better as a proof in my books. As it approach zero in the two paths. Hence the limit is continuous for $(0,0)$

Now i know the definition of continuity is formally: A function f is continuous at a point $a$ if for every $\epsilon>0$, there exists $\delta>0$ such that $|x−a|<\delta$ implies that $|f(x)−f(a)|<\epsilon$

But i get confused at finding $\delta$ and $\epsilon$

So what i usually use is $x=a$ $$\lim_{x \to a} f(x) = f(a)$$

I'm guessing this is not enough proof as we have not proven that $f(x)$ continuous along every point of the domain.

• In the first limit you cannot say $\frac{0}{0} = 0$. A more proper way to write this is "along $x = y$, we have $\frac{2x^2y}{x^4+y^2} = \frac{2x^3}{x^4+x^2} \to 0$ as $x \to 0$". – Empiricist Aug 26 '15 at 23:57
• Oh ok so me taking two different paths in the following two limit is wrong. Also did you mean to say y=ax? such that $\frac{2ax^3}{x^4+a^2x^2}$ – Patrick Aug 27 '15 at 0:27
• Hmm.. I wrote $x = y$ simply because you wrote $\lim_{x=y\to 0}$ in the question. Analyzing the behavior along $y = ax$ is even better, but it may cause confusion as you also mentioned the continuity at the point $a$ later in the question. – Empiricist Aug 27 '15 at 2:05
• It is not continuous at $(0,0)$; all straight-line paths converge to $0$, but some curved paths don't. – Akiva Weinberger Aug 27 '15 at 2:15

Here's a hint: what if you approach $(0,0)$ along the path $y = x^2$? Note that $f$ is only continuous at $(0,0)$ if every path to $(0,0)$ yields the same limit.

• $y=x^2$ but we're talking about $$\frac{2x^2y}{x^4+y^2}$$ did you just take out $x^2$ from that? – Patrick Aug 26 '15 at 23:57
• No. I'm saying instead of using the line $y=ax$, try the parabola $y=x^2$ as your path to get to $(0,0)$. – Cameron Williams Aug 26 '15 at 23:57
• or take $y=-{ x }^{ 2 }$ – haqnatural Aug 27 '15 at 0:11
• Ahhh... f(x) = 1 for all values along $y=x^2$ the limit for all values is 1. So for point x=0 the limit does not equal f(0) = 1 – Patrick Aug 27 '15 at 0:31
• @Patrick Exactly right. – Cameron Williams Aug 27 '15 at 0:47

$$\frac{2x^2y}{x^4+y^2}$$

I order for a limit to exist it must so follow that at (0,0) we get the same limit among all paths. The limit 0 for f(x,y) along y=ax approaching (0,0) is only one direction. It must follow through for other directions.

A long the path $y=ax$

$$\lim_{x \to 0} f(x,ax) = lim_{x \to 0} \frac{2ax^3}{2x^2(x^2+a)} = lim_{x \to 0} \frac{2ax}{x^2+a} =0$$

As give. Now for a function to be continous for x=a, i must so follow that. $$\lim_{x \to a} f(x) = f(a)$$

$$f(0, ax) = \frac{2a*0}{0+a} = 0$$

A long the path $y=x^2$ $$f(x, x^2) = \frac{2x^4}{2x^4} = 1$$ $$f(x, x^2) = 1 \quad \text{for all value}$$

Hence $$\lim_{x \to 0} f(x, x^2) = 1$$

$\therefore$ After looking at another path we find that the Limit actually doesn't exist as we get different limits from different paths at point (0,0).