If the function $f$ is defined on an unbounded above domain $D \subseteq \Re $ and is eventually monotone and eventually bounded, then $ \lim_{x\rightarrow \infty} f(x)$ is finite

I tried to workout the proof as:

Since $f$ is eventually monotone $\Rightarrow \exists x^*, x^* \leq x_1 < x_2 $ we have $f(x_1) \leq f(x_2)$

and since $f$ is eventually bounded $\Rightarrow \exists \hat{x},\ \exists \ L \leq M \in \mathbb{R} \ s.t. \lim_{x\rightarrow \infty} f(x) = L \\\forall \ \hat{x}\leq x $

Take $x = max(\hat{x}, x^*)$ and we have $\lim_{x\rightarrow \infty} f(x) = L$

  • $\begingroup$ I need critique and suggestions please $\endgroup$ – Logarithm Oct 8 '13 at 1:57

First, because $f$ is eventually monotone (without loss of generality increasing), you know that there is an $x^*$ such that for all $x^*\leq x_1\leq x_2$ you have $f(x_1)\leq f(x_2)$. I'm not sure why you've got $\forall x_1,x_2\in D$. Now, since $f$ is eventually bounded, you have that there exists an $M$ and $\hat{x}$ such that for all $x\geq \hat{x}$, $f(x)\leq M$. Take $x'=\operatorname{max}(x^*,\hat{x})$. Then $f$ is bounded and monotone (increasing) on $(x',\infty)$. This implies it has a limit.

Your "eventually bounded" part is a little confused. And, I'm not getting what you are saying.

  • $\begingroup$ Johnson: I have $\forall x_1,x_2 \in \ D$ to specify that they belong to the domain. I think you are right about the boundedness. I need to fix it as you stated. $\endgroup$ – Logarithm Oct 8 '13 at 2:15
  • $\begingroup$ @Logarithm By using $\forall$ it seems that you are saying that $f(x_1)\leq f(x_2)$ for all elements of the domain. $\endgroup$ – Joe Johnson 126 Oct 8 '13 at 2:16
  • $\begingroup$ @ Joe Johnson: I see what you mean now $\endgroup$ – Logarithm Oct 8 '13 at 2:18
  • $\begingroup$ Johnson: I think because I was reading the definition of a definition of 'eventually increasing sequence'. I don't know why when I type your full name with @ in front of it came out as 'Johnson' $\endgroup$ – Logarithm Oct 8 '13 at 2:22
  • 1
    $\begingroup$ @Logarithm When you reply to my answer, there is no need to put the @ symbol. It will still appear in my inbox. So, it doesn't automatically complete it to JoeJohnson 126. $\endgroup$ – Joe Johnson 126 Oct 8 '13 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.