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

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.)

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.

• 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). – Semiclassical Jan 8 '18 at 14:35
• @Rick I am not sure what you are suggesting. Are you suggesting, as others have, that using non-straight paths will disprove the limit? – badjohn Jan 8 '18 at 16:01

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.

• 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 $(0,0)$. – badjohn Jan 8 '18 at 15:58
• @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)? – A.Γ. Jan 8 '18 at 16:40
• 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. – badjohn Jan 8 '18 at 17:42

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!)

• 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. – badjohn Jan 8 '18 at 15:51

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 : $$f(x_n,y_n)=\Big(\frac{1}{n^4}+\frac{1}{n} \Big)\tfrac{1}{\sqrt{n}}\times n^2 = \frac{1}{n^2\sqrt{n}}+\sqrt{n}$$

• 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. – badjohn Jan 8 '18 at 15:50