Limit of $\frac{4x^2y}{x^2-y^2}$ I am wondering what is the limit of $\frac{4x^2y}{x^2-y^2}$ when $(x,y)$ approaches $(0,0)$. If I use the polar coordinates it is obvious that
$$\frac{4x^2y}{x^2-y^2} = r\cdot\frac{2\sin(2\theta)\cdot \cos(\theta)}{\cos(2\theta)} = r\tan(2\theta)\cos(\theta) $$ If we are approaching $(0,0)$ from any direction $\theta \neq \frac{\pi}{4}$ then of course the limit is $0$ as $r \rightarrow 0$ but can I say the same when $\theta \rightarrow \frac{\pi}{4}$. Should I even consider this case as it is not in a domain o function?
Thanks for any help.
 A: The pair of lines $\{(x,y)\,:\, x=y\lor x=-y\}$ is not in the domain of the function, so you should not consider it (as long as you've ascertained that $(0,0)$ is an accumulation point of the domain). However, in the domain there are sequences $p_n\to 0$ such that $f(p_n)\not\to 0$ as $n\to\infty$, more or less because of the fact that $\lvert\tan(2\theta)\cos\theta\rvert\to \infty$ as $\theta\to\frac\pi4$. For each $r>0$, you can find some $\theta_r\notin\{\frac\pi4,\frac34\pi,\frac 54\pi,\frac74\pi\}$ such that $\lvert\tan(2\theta_r)\cos\theta_r\rvert\ge\frac1r$. Therefore, if you call $(x_r,y_r)=(r\cos\theta_r,r\sin\theta_r)$, you obtain that $(x_r,y_r)\to (0,0)$ as $r\to 0$ and $\limsup\limits_{r\to 0}\lvert f(x_r,y_r)\rvert\ge1$.
A: If you approach the origin along the curve $y=x+x^2$, you get (after a short calculation) the limit $-2$, which is different from the zero that you got along straight lines, and this shows that the multivariable limit doesn't exist. What happens is that even if $\theta$ is never equal to $\pi/4$ (which is of course forbidden), it depends on $r$ in such a way that it tends to $\pi/4$ as $r\to 0$, and the factor that you have together with the factor $r$ becomes unbounded.
For the argument with polar coordinates to work, you need a “zero-approaching” function of $r$ times a bounded factor. Here and
here
are a some related answers that I've written before about that.
A: As noticed, points such that $\theta = \frac{\pi}{4}$ are not included in the domain but it doesn't mean we can't approch them in such way the expression diverges.
In this case the key point is that by polar coordinates we obtain an expression in the form $r\cdot f(\theta)$ with $f(\theta)$ not bounded and this is a strong clue that limit doesn't exist.
Indeed we can see that by $y=0 \implies \frac{4x^2y}{x^2-y^2}=0$ but for $x=t\to 0$ and $y=t- t^2$ we obtain
$$\frac{4x^2y}{x^2-y^2}=4\frac{t^3-t^4}{2t^3-t^4}=4\frac{1-t}{2-t} \to 2$$
and for $x=t\to 0^+$ and $y=t- t^3$ we obtain
$$\frac{4x^2y}{x^2-y^2}=4\frac{t^3-t^5}{2t^4-t^6}=4\frac{1-t}{2t-t^3} \to \infty$$
which is indeed related to the not bounded function $f(\theta)$ you have found in the polar coordinates expression.
