Computing the limit at infinity for multivariable function

Question

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.

Answer 1

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 $(\infty,\infty)$.

Answer 2

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 $y=0$

$$xe^{-y}= te^0=t\to \infty $$