Computing the limit at infinity for multivariable function I seem to be having problems understanding the epsilon-N definition of limits when the function takes multiple variables.
For example, consider the limit $\lim_{(x,y) \rightarrow (\infty, \infty)} xe^{-y}$, which has come up in my stats homework. My hunch is that this limit should converge to $0$, as this yields the correct answer and the graph seems to "flatten out" in general when looking far away in the first quadrant.
Yet, I can neither confirm nor disprove this guess since I cannot find the definition of limits of multivariable functions at infinity. The only definition I could find are those at finite points, in which case a direct generalization of $\epsilon-\delta$ definition for single variable functions could be applied.
Could somebody please explain the rigorous definition of limits at infinity? Also, if possible, could you confirm or disprove my guess about $\lim_{(x,y) \rightarrow (\infty, \infty)} xe^{-y}$?
Thanks very much.
 A: In this case, the limit is not well-defined.  Specifically, it depends on the path you take to get to $(\infty, \infty)$.  For example, if you fix $x$ and take $y$ to $\infty$, you will see that the function goes to zero everywhere.  If you then take $x$ to infinity, well zero stays zero.
If you do it in the opposite order (fix $y$ and take $x$ to $\infty$, then take $y$ to $\infty$), you will get that the function blows up.
In general, multivariate functions -- even nice continuous, smooth ones like $xe^{-y}$ -- will not have good limits as you go to infinity.  You would need another property (like uniform convergence) to talk about the limit as you go to $(\infty,\infty)$.
A: Taking this limit we are considering paths for which $\|(x,y)\|=\sqrt{x^2+y^2}\to \infty$ and in this case limit doesn't exist indeed as noticed in the comments

*

*for $x=y=t\to \infty$
$$xe^{-y}= \frac t {e^t} \to 0$$
but

*

*for $x=t\to \infty$ and $y=\log t \to \infty$
$$xe^{-y}= \frac {t} {e^{\log t}}=\frac t t=1 $$
or also

*

*for $x=t\to \infty$ and $y=0$
$$xe^{-y}= te^0=t\to \infty $$
