$f(x,y)=\begin{cases} |\frac{y}{x^2}|exp(-|\frac{y}{x^2}|) & , x\ne0\\ 0 & , x=0 \end{cases}$ $f(x,y)=\begin{cases}
|\frac{y}{x^2}|exp(-|\frac{y}{x^2}|) & , x\ne0\\
0 & , x=0
\end{cases}$
I need to show that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ the limit does not exist in $(0,0)^T$. I tried to prove it with the sequence criteria but I could not find a good sequence. Polar form doesn't get me further. 
 A: On $y=x^2$ you have
$$\lim\limits_{(x,y)\to(0,0)}\left|\frac{y}{x^2}\right|\exp\left(-\left|\frac{y}{x^2}\right|\right)
=\lim\limits_{(x,y)\to(0,0)}\exp(-1)=\frac{1}{e}$$
Denote
$$f(x,y)=\left|\frac{y}{x^2}\right|\exp(-\left|\frac{y}{x^2}\right|)
$$
on $y=0$
 we have $f(x,0)=0$
and therefore 
$$\lim\limits_{(x,y)\to(0,0)}f(x,y)=
\lim\limits_{x\to0}f(x,0)=0$$
thus, the limit doesn't exists because otherwise it would be the same on every path\curve.
A: Choose two paths:
1) $y=e^x-1$. Then $\lim_{(x,y)\to 0^+}f(x,y)=\lim_{x\to0^+}\left|\frac{e^x-1}{x^2}\right|\exp(-\frac{e^x-1}{x^2})=\lim_{x\to 0^+}\left|\frac{1}{x}\right|\exp(-\frac{1}{x})=0$
This is a path for which $y$ reaches zero at the same pace as $x$. You can convince yourself that the limit is then $0$.
2) $y=x^2$, for the which the limit is easily seen to be $e^{-1}$.
As the limit depends on how we take it (the path we folow) it is not well defined.
Therefore, the limit at $(0,0)$ does not exist for this function.
(A third alternative path could be taking the one in 1) and considering the limit as $x\to0^-$. In this case the exponential diverges as well as the multiplicative factor and, thus, so does the limit as well. EDIT: I forgot take the absolute value of the exponent. Hence, this third path won't apply to the function in the problem statement. The other two are still valid though.)
