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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!

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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
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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $m$

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

So

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

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  • $\begingroup$ 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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