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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.

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    $\begingroup$ 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. $\endgroup$
    – Ninad Munshi
    Commented Sep 28, 2021 at 14:38
  • $\begingroup$ Let $(x_n,y_n) = (n, log(n))$. Then $(x_n, y_n) \to (\infty, \infty)$, but $f(x_n, y_n) = n/n = 1$. $\endgroup$
    – user3716267
    Commented Sep 28, 2021 at 14:39
  • $\begingroup$ 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? $\endgroup$
    – 이희원
    Commented Sep 28, 2021 at 15:01

2 Answers 2

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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 $(\infty,\infty)$.

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  • $\begingroup$ Huh! Thanks for the counterexample! Is there a foolproof definition of limits at infinity, just like the epsilon-N for single variable functions? $\endgroup$
    – 이희원
    Commented Sep 28, 2021 at 14:58
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    $\begingroup$ 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. $\endgroup$
    – Craig
    Commented Sep 28, 2021 at 15:41
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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 $y=0$

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

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