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Problem

if we have function defined as:

$$ F(x,y)=\frac{xy}{1+e^{x-y}} \quad \text{when } (x,y)\in\mathbb{R}^2 $$

in which points function is continuous ?

Attempt to solve

Just by looking at function we can see that function should be defined $\forall \text{ }\mathbb{R}$ when denominator $(1+e^{x-y})\neq 0$. Another possible case where continuity is not guaranteed is when $\frac{0}{0}$ meaning situation where $F(0,0)$

We could try to found limit at $F(0,0)$ $$\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{1+e^{x-y}}=0$$

It would seem function behaves nicely close to the $0$. Now i don't know how to proof that vector valued function has limit but i do know how to prove when function does not have limit at given point. If we can approach the limit in two different ways and resulting in two different limits. Then we can conclude that no limit exists because of this.

I haven't been able to prove this function doesn't have limit at point $(0,0)$. It would be possible that limit exists in $(0,0)$ but how do you prove this.


If someone knows how to prove function has limit in given point that would be highly appreciated.

Also i've tried to understand the concept of $\delta, \epsilon$ definition/proof for limits. I think i have some level of understanding on how limits can be defined with $\delta,\epsilon$ but unable to make a proof using this.

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    $\begingroup$ composition, sum and product of continuous functions is continuous. If $f$ is continuous, then $\frac{1}{f}$ is continuous away from $f=0$. Also, note, $F(0,0)$ is not $\frac{0}{0}$, it is $\frac{0}{2}=0$ $\endgroup$ – Callus Jan 13 '18 at 23:56
  • $\begingroup$ @Callus yes denominator in this function cannot be $0$ with real value input. $\endgroup$ – Tuki Jan 13 '18 at 23:58
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If you have already proven with the $\delta,\epsilon$ definition that sums, products, quotients (where the denominator is non-zero) and compositions of continuous functions (and the projections $x,y$ of course) are continuous, then you can directly say that $xy$, $x-y$,$1+e^t$, $\frac{1}{1+e^t}$ (because $1+e^t>0$) are and thus is $F(x,y)$ continuous on $\mathbb{R}\times\mathbb{R}$.

If you want to calculate the limit, you can just estimate $$\lim_{(x,y)\rightarrow(0,0)}\left|\frac{xy}{1+e^{x-y}}\right|\leq\lim_{(x,y)\rightarrow(0,0)}|x||y|=0$$ because $1+e^{x-y}>1$, so we have $\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{1+e^{x-y}}=0$.

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