TITLE: Does $\frac{(x^2 + y^2) y}{x}$ have a limit at $(0,0)$?
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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!)