4
$\begingroup$

Does $\frac{(x^2 + y^2) y}{x}$ have a limit at $(0,0)$?

Recently, someone asked whether a function from $\mathbb{R}^2$ to $\mathbb{R}$ had a limit at $(0,0)$. The question was easy and answered in the negative by showing that approaching $(0,0)$ on different lines led to different limits.

This prompted a question: is there such a function which has a limit when restricted to any straight line through $(0,0)$ and the limit is the same in all cases yet the function does not have a limit at $(0,0)$?

This led me to consider this function:

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

This looks a bit nicer in polar coordinates with $x = r \sin \theta$ and $y = r \cos \theta$

$$ f(x, y) = \begin{cases} r^2 \tan \theta, & \text{if $\theta \neq \pm \frac{\pi}{2} $} \\ 0, & \text{if $\theta = \pm \frac{\pi}{2} $} \end{cases} $$

So, if the function is restricted to a straight line through $(0,0)$ then the function clearly has the limit $0$ since $\tan \theta$ will be a constant.

However, it is not continuous at $(0,0)$ as within any radius of $(0,0)$, it takes arbitrarily large values.

So, here is my question: is the above right or have I made a mistake? (I am rather rusty in this area.)

Additional clarification

I know that I don't need to restrict myself to straight lines when testing limits. In fact, that was the point of the exercise: to show that straight lines may disprove a limit but testing only straight lines will not prove a limit. I wanted an example that had a limit along all straight lines yet still failed to have a limit.

Simpler examples that demonstrate this would be welcome.

$\endgroup$
  • $\begingroup$ You could also consider how the function behaves on a path like $y=ax^r$ and see if there’s any pair of parameters which don’t give a limit of zero. (This isn’t enough to confirm continuity, but it can be used to disprove it). $\endgroup$ – Semiclassical Jan 8 '18 at 14:35
  • $\begingroup$ @Rick I am not sure what you are suggesting. Are you suggesting, as others have, that using non-straight paths will disprove the limit? $\endgroup$ – badjohn Jan 8 '18 at 16:01
3
$\begingroup$

Hint: try the limit along $y=x^{1/3}$. Is it also zero?

The easiest way is to rewrite $$ \frac{(x^2+y^2)y}{x}=xy+\frac{y^3}{x}. $$ The first term goes to zero, so you need to study the second term only.

$\endgroup$
  • 1
    $\begingroup$ Thanks. The motivation for my definition was to get the $\tan \theta$ term in the polar for as this would mean that the function attained arbitrarily large values on any circle around $(0,0)$. $\endgroup$ – badjohn Jan 8 '18 at 15:58
  • $\begingroup$ @badjohn $\frac{y^2}{x}$ would probably be an easier example. Though the fact that you have to additionally define the function along the whole $y$-axis is a bit tense. Maybe $\frac{x^2y}{x^4+y^2}$ instead (and zero at the origin)? $\endgroup$ – A.Γ. Jan 8 '18 at 16:40
  • 1
    $\begingroup$ Thanks. I mentioned my motivation for my function but I failed to consider simplifying it after finding it. Actually, I had $\frac{y}{x} \sqrt{x^2 + y^2}$ first as I was considering $r \tan \theta$ in polar form. My function above was a slight simplification of that as the square root was not required. $\endgroup$ – badjohn Jan 8 '18 at 17:42
2
$\begingroup$

What you wrote is correct.

Another way to see this function doesn't have a limit is to approach along the $y = x^\frac{1}{4}$ curve. (You don't have to limit yourself to approaching along a straight line!)

$\endgroup$
  • 1
    $\begingroup$ I know that I don't have to restrict myself to a straight line. That was actually the point of the exercise: to show that although the straight line approach will often disprove a limit, it cannot be used to prove a limit. $\endgroup$ – badjohn Jan 8 '18 at 15:51
1
$\begingroup$

No, it does not have a limit at $(0,0)$ : you just have to use the sequence $(x_n,y_n):=(\tfrac{1}{n^2},\tfrac{1}{\sqrt{n}})$ to see it : $$f(x_n,y_n)=\Big(\frac{1}{n^4}+\frac{1}{n} \Big)\tfrac{1}{\sqrt{n}}\times n^2 = \frac{1}{n^2\sqrt{n}}+\sqrt{n}$$

$\endgroup$
  • $\begingroup$ Thanks. I did not calculate the coordinates of large valued points near the origin because they clearly existed due to the $\tan \theta$ term. I could pick a circle around $(0,0)$ as small as I like but by picking a suitable $\theta$, I could make the value of $f$ as large as I liked. $\endgroup$ – badjohn Jan 8 '18 at 15:50

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.