I'm trying to show that the function $f: \mathbb{R^2} \rightarrow \mathbb{R}$ given by

$ f(x,y)= \begin{cases} \frac{xy}{x^2+y^2}, \text{ if } (x,y) \neq (0,0) \\ 0 \text{ if } (x,y) = (0,0) \\ \end{cases} $

is continuous along every horizontal and every vertical line. From what I understand, I need to show that for every $x_0, y_0 \in \mathbb{R}$, the functions $g, h: \mathbb{R} \rightarrow {R}$ given by $g(t) = f(x_0, t)$ and $h(t) = f(x_0, t)$ are continuous.

I also need to show that the function is not continuous on (0,0).

From the definition, I understand that $f$ is continuous at $p \in E$ if $\forall \epsilon > 0$, $\exists \delta >0$ such that $\forall x \in E$, $d(x,p) < \delta$ implies $d(f(x), d(p)) < \epsilon$.

Here's what I have so far on trying to prove continuity:

For horizontal line, we can set $y=c$ and so the function gives us $f(x,c) = \frac{xc}{x^2 + c^2}$. Similarly for the vertical line, we have $x=c$ and so $f(c,y) = \frac{cy}{c^2 + y^2}$.

But I'm really confused on how I should prove that the function is continuous along the vertical and horizontal lines.

As for proving not continuous at (0,0), can I just say that the function is not defined on (0,0)?

Help would be much appreciated!


marked as duplicate by Arnaud D., Claude Leibovici, Lord Shark the Unknown, Cesareo, Deepesh Meena Sep 20 '18 at 18:38

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  • $\begingroup$ Hint: You can use polar coordinates $x=\rho\cos\theta$ and $y=\rho\sin\theta$ $\endgroup$ – MattG88 Nov 14 '16 at 2:34
  • $\begingroup$ What's the limit of the single variable functions you pointed out? What do those limits say about the continuity? $\endgroup$ – Kaynex Nov 14 '16 at 2:45
  • $\begingroup$ @Kaynex Ok, so I went through Rudin and I found that f is continous iff $ lim_{x\rightarrow p} f(x) = f(p)$. And so $lim_{x\rightarrow p} = \frac{pc}{p^2 + c^2} = f(p)$? And so we have continuity? $\endgroup$ – Nikitau Nov 14 '16 at 3:00

It should be clear that the function is continuous everywhere that is not the origin. So on all vertical and horizontal lines that are not the $x$-axis or the $y$-axis, their continuity is clear.

Now on the x axis and y axis your function is the constant zero function which is obviously continuous.

To show that it is not continuous at zero, the easiest way is to find a line through the origin so that if you approach $(0,0)$ on it, the limit is non-zero.

  • $\begingroup$ Thanks for the reply! Intuitively, I understand why the function is continuous everywhere except the origin. I think Rudin's definition is really throwing me off -- how exactly do I prove continuity? How do I find a $\delta$ such that $d(x,p) < \delta$ implies $d(f(x), d(p)) < \epsilon$? $\endgroup$ – Nikitau Nov 14 '16 at 2:43

If you use polar coordinates (see the Hint above)

\begin{equation} f(x,y)=\cos\theta\sin\theta \end{equation}

so if you evaluate the limit for $\rho\rightarrow 0$

\begin{equation} \lim_{\rho\rightarrow 0}\frac{\rho^2\cos\theta\sin\theta}{\rho^2(\cos^2\theta+\sin^2\theta)}=\lim_{\rho\rightarrow 0}\cos\theta\sin\theta=\cos\theta\sin\theta \end{equation}

it depends by $\theta$ i.d. $f(x,y)$ is not continuous on zero.

  • $\begingroup$ This is probably a really stupid question, but are we evaluating the limit at $\rho \rightarrow 0$ because we're determining continuity at the point (0,0)? I'm still trying to get used to the definitions. Thanks! $\endgroup$ – Nikitau Nov 14 '16 at 3:03
  • $\begingroup$ Oh yes!! I've overlook it.. $\endgroup$ – MattG88 Nov 14 '16 at 3:05
  • $\begingroup$ Not a problem! :) Like I said, I'm just trying to wrap my head around the definitions. Thank you! $\endgroup$ – Nikitau Nov 14 '16 at 3:07

Let us fix $y=k$ (along horizontal line)

We take $k\neq 0$ otherwise $f(x)=0$ which is continuous automatically.

Then the function is given by $f(x)=\dfrac{kx}{x^2+k^2};x\neq0$and $f(x)=0;x=0$

Then $\lim_{x\to 0}f(x)=0=f(0)$. Hence $f$ is continuous along every horizontal line.

Similarly take $x=k$( along vertical line). Repeat the same arguments.

For showing discontinuity;Use y=mx where $y\to 0$ as $x\to 0$.

Then $f(x,mx)=\dfrac{m}{m^2+1}$ which tends to different values for different values of $m$

  • $\begingroup$ Thank you for the reply! This might be a dumb question, but I'm having a lot of difficulty in understanding and applying Rudin's theorems. So when Rudin says that to prove continuity, we need $lim_{x\rightarrow p} f(x) = f(p)$, how do we know what to plug for p? I know p is the point of convergence, but how do we determine it before proving convergence? $\endgroup$ – Nikitau Nov 14 '16 at 3:19
  • $\begingroup$ Well you can do like this.. Take $x=p+h$ where $h\to 0$ See what is the definition of the function when $x>p$ ;plug in there and then put $h=0$ in the expression $\endgroup$ – Learnmore Nov 14 '16 at 4:06

Take $\gamma_{1}(t)=(0,t)$ and $\gamma_{2}(t)=(t,t)$.


$\lim\limits_{t\to 0} f(\gamma_{1}(t))=compute=0$ and $\lim\limits_{t\to 0} f(\gamma_{2}(t))=compute=\frac{1}{2}$.

So it does not exist $\lim\limits_{(x,y)\to (0,0)}f(x,y) \Rightarrow$ $f$ is not continuous in $(0,0)$.

To see what f is continuous along every horizontal, just note that $\lim\limits_{(x, y_{0})\to (a\not=0,y_{0})}f(x,y)=f(a,y_{0})$. Similarly, it is concluded that is continuous along every vertical line.

p.s. On horizontal continuity, we may have to $a=0$, in this case $y_{0}\not=0$. This observation is analogous for the analysis of continuity (Calculation of the limit) on the vertical axis.

  • $\begingroup$ Does the limit not exist because limits of functions have to be unique? Sorry if this is a dumb question. $\endgroup$ – Nikitau Nov 14 '16 at 3:09
  • $\begingroup$ Please do not apologize. We are all learning! Yes, the limit when it exists is unique! $\endgroup$ – Manoel Nov 14 '16 at 3:12

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