# Prove that if $f$ is eventually monotone and eventually bounded $\Rightarrow \lim_{x\rightarrow \infty} f(x)$ is finite

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 $\lim_{x\rightarrow \infty} f(x) = L$

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I need critique and suggestions please –  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.
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. –  Logarithm Oct 8 '13 at 2:15
@Logarithm By using $\forall$ it seems that you are saying that $f(x_1)\leq f(x_2)$ for all elements of the domain. –  Joe Johnson 126 Oct 8 '13 at 2:16