Prove non-existance of limit: $f(x,y) = \frac{xy\sin(\frac{x}{y})}{x^2 + |y|^3}$ I need to prove that $f(x,y) = \frac{xy^2\sin(\frac{x}{y})}{x^2 + |y|^3}$ does not tend to $0$ when $(x,y)$ approaches $(0,0)$.
In order to do so, I would need to find some direction $\alpha$ such that $f(\alpha(t))$ approaches to some value $L \neq 0$ as $(x,y)$ approaches the origin.
The problem is that I can't seem to find that direction. What could I try?
I found  $$f(x^\frac{1}{2}, x^\frac{1}{3}) = \frac{x^\frac{7}{6}  \sin(x^\frac{-1}{6})}{2x} = \frac{x^\frac{1}{6}  \sin(x^\frac{-1}{6})}{2} = \frac{\sin(x^\frac{-1}{6})}{2 x^\frac{-1}{6}} \rightarrow \frac{1}{2}$$
Is this correct?
 A: Using the path $(kt,t)$ as $t\to 0$ and assuming $k>0$
$$L=\lim_{t \to 0}\frac{kt.t \sin k}{k^2t^2+|t|^3}$$
$$L=\lim_{t \to 0}\frac{k\sin k}{k^2+\frac{|t|^3}{t^2}}$$
Whether you approach it from positive or negative side the answer will be 
$$L=\frac{\sin k}{k}$$
Which is dependant on the path that is $k$.
Which shows that there's no unique value of $L$ hence your limit doesn't exist. 
Note that your limit is correct but only showing that limit across a fixed path exist and $L\neq 0$ doesn't mean it does exist or not. You have to show it's independent of path aswell.
A: Used polar coordinates $x=r\cos\phi, y=r\sin\phi$ and compute the limit $r \rightarrow 0$. The result is
\begin{align}
\lim_{r\rightarrow 0} \frac{r^2\cos\phi\sin\phi \cdot \sin(\cot\phi)}{r^2 \cos^2\phi + |r\sin\phi|^3} 
&= 
\lim_{r\rightarrow 0} \frac{\cos\phi\sin\phi \cdot \sin(\cot\phi)}{\cos^2\phi + r|\sin^3\phi|} 
\\ &=
\frac{\cos\phi\sin\phi \cdot \sin(\cot\phi)}{\cos^2\phi} 
\\ &= \tan\phi \cdot \sin(\cot\phi)
\end{align}
which is a nonconstant function of $\phi$, thus your limit does not exist.
EDIT: The answer of Mann is easier and more straightforward though.
