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


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.

  • $\begingroup$ 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 $x \to 0$". $\endgroup$ – Empiricist Aug 26 '15 at 23:57
  • $\begingroup$ 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 $\frac{2ax^3}{x^4+a^2x^2}$ $\endgroup$ – Patrick Aug 27 '15 at 0:27
  • $\begingroup$ 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. $\endgroup$ – Empiricist Aug 27 '15 at 2:05
  • $\begingroup$ It is not continuous at $(0,0)$; all straight-line paths converge to $0$, but some curved paths don't. $\endgroup$ – 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.

  • $\begingroup$ $y=x^2$ but we're talking about $$\frac{2x^2y}{x^4+y^2}$$ did you just take out $x^2$ from that? $\endgroup$ – Patrick Aug 26 '15 at 23:57
  • 1
    $\begingroup$ No. I'm saying instead of using the line $y=ax$, try the parabola $y=x^2$ as your path to get to $(0,0)$. $\endgroup$ – Cameron Williams Aug 26 '15 at 23:57
  • 1
    $\begingroup$ or take $y=-{ x }^{ 2 }$ $\endgroup$ – haqnatural Aug 27 '15 at 0:11
  • $\begingroup$ 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 $\endgroup$ – Patrick Aug 27 '15 at 0:31
  • 1
    $\begingroup$ @Patrick Exactly right. $\endgroup$ – Cameron Williams Aug 27 '15 at 0:47


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.