# Prove that $f(x,y)$ is continuous in $(0,0)$

Prove that $f(x,y)$ is continuous in $(0,0)$, where

\begin{equation} f(x,y) = \begin{cases} \frac{x^2y}{x^4+y^2}, & (x,y)\neq 0\\ 0, & (x,y) = (0,0) \end{cases} \end{equation}

The solution I have is that f is not continuous in $(0,0)$. (The solution doesn't say more than that.)

However, the result I got is that $f$ is continuous in $(0,0)$. Here's my approach:

Lets transform $x$ and $y$ into their polar coordinates, so that we can approach $(0,0)$ from any direction by varying $\theta$:

$(x,y) = (r\cos(\theta),r\sin(\theta)) \qquad r\in\mathbb{R}^+_0 \quad \theta\in[0,2\pi)$

Then $f$ is continuous iff

\begin{equation} \lim_{(x,y)\to (0,0)} f(x,y) = 0 = f(0,0) \end{equation}

By using the polar coordinates and letting $r\to 0$ we get:

\begin{equation} \lim_{(x,y)\to (0,0)} f(x,y) = \lim_{r\to 0} \frac{r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta + r^2\sin^2\theta} = \lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta} \end{equation}

By a case distinction by $\theta$ we get:

1. $\theta\in[0,2\pi)\backslash\{0,\pi\}$: \begin{equation} \lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta} = \lim_{r\to 0} \frac{0}{\sin^2\theta} = 0 = f(0,0) \end{equation}
2. $\theta\in\{0,\pi\}$: $\Longrightarrow \sin\theta = 0$ \begin{equation} \lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta} = \lim_{r\to 0} \frac{0}{r^2\cos^4\theta} = 0 = f(0,0) \end{equation}

Form 1. and 2. we can conclude that $f$ is continuous in $(0,0)$.

What am I doing wrong in my approach?

• As you have to prove this in $\Bbb{R^2}$, using this way you can only disprove continuity. You have to use the definition of limit in order to prove that limit is indeed 0. You calculated the limit along two paths doesn't mean that the limit is actually 0 – JukesOnYou May 23 '15 at 16:53
• The function isn't continuous. Take the path $t\mapsto (t,t^2)$. – Git Gud May 23 '15 at 17:34

You have shown the limit is $0$ along all straight lines through $(0,0).$ That's not enough: There are tons of paths to $(0,0)$ that are not straight lines. For example, the path $(x,x^2)$ as $x\to 0.$

• I see, this would make the limit be $\frac{1}{2}$ and thus not 0. Which means that $f$ is not continuous. – ndrizza May 23 '15 at 18:54

What you're doing wrong is that you're separating both $r$ and $\theta$ variables. This is not following the continuity definition for functions with several variables.

Please recall the definition of continuity for a function of several variables.

• I see, thank you. By coincidence Wikipedia has just my question as an example: en.wikipedia.org/wiki/… – ndrizza May 23 '15 at 18:52

So to check for continuity you need to determine that the limit exists, i.e. check all sides, and then check that it is equal to the function. So you didn't quite do that correctly. You need to check the limit as $x$ approaches $0$ say along $y^2$ for example. All you have to do is find one case where the limit does not exist and thus it is not equal from all sides. Therefore the limit does not equal the function at that point and it is discontinuous.

So just look for one case where the limit does not exist. Check this link out if you don't remember that:

http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx

• I don't have high enough reputation to upvote it but the comment above makes an excellent observation. – Aaron Zolotor May 23 '15 at 16:57
• Since I'm saying that $\theta$ may be any angle $\theta\in[0,2\pi)$ I have the impression that I'm taking into account all directions to approach $(0,0)$. Or am I wrong with that? – ndrizza May 23 '15 at 17:01
• See the comment on the continuity of variables. – Aaron Zolotor May 23 '15 at 17:09