# 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.

• The general definition for multivariate limits is that they must exist along all paths. However, consider the path $x=e^y$ which goes to $(\infty,\infty)$, but the limit approaches $1$. The path $x=y$ goes to $0$ - two different paths yielding two different limits means the limit doesn't exist. Sep 28 at 14:38
• Let $(x_n,y_n) = (n, log(n))$. Then $(x_n, y_n) \to (\infty, \infty)$, but $f(x_n, y_n) = n/n = 1$. Sep 28 at 14:39
• Thanks for the counterexample! I did consider the path-based definition, but it seemed a little bit problematic, because when I was introduced to the epsilon-delta definition for multivariable functions, I was told that it is preferable to define this new definition as a generalization, rather than a consequence of single variable limits. Is there another definition that does not use the single variable limits?
– 이희원
Sep 28 at 15:01

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 $$(\infty,\infty)$$.

• Huh! Thanks for the counterexample! Is there a foolproof definition of limits at infinity, just like the epsilon-N for single variable functions?
– 이희원
Sep 28 at 14:58
• It is well-defined (limit = C) iff you can say $\forall \epsilon >0 \exists M,N s.t. \forall x > M \forall y > N |f(x,y) - C| < \epsilon$. The fundamental idea is, given an $\epsilon$, can you find an open set containing your limit point where the function is less than $\epsilon$ away from your limit value. Sep 28 at 15:41

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 $$y=0$$

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