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Prove that if $f\colon[a,\infty)\longrightarrow R$, $f$ monotonically increasing and $\lim_{x\to \infty} f(x)$ exists then it is also upper bounded

How do I prove this?

The only thing that I could find was this but it requires to prove that limit exists given that f is already bounded above and increasing.

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  • $\begingroup$ Can you draw a picture of what this kind of $f$ would look like? After drawing those pictures, do you see a candidate for what an upper bound on $f$ could be? $\endgroup$ – angryavian Nov 7 '17 at 17:40
  • $\begingroup$ Well, it would certainly increase. But how can I assume the upper bound? There is no info about the value of the limit $\endgroup$ – Dimitris Moraitidis Nov 7 '17 at 17:44
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Let $M=\lim_{x\to+\infty}f(x)$. Could there be a $x_0\in[a,+\infty)$ such that $f(x_0)>M$? No, because $f$ is monotonically increasing and\begin{align}f(x_0)>M&\implies\bigl(\forall x\in[x_0,+\infty)\bigr):f(x)\geqslant f(x_0)>M\\&\implies\lim_{x\to+\infty}f(x)\geqslant f(x_0)>M,\end{align}which is a contradiction, since $M=\lim_{x\to+\infty}f(x)$.

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  • $\begingroup$ Thanks a lot. The clue was after all to assume that value of M is finite number $\endgroup$ – Dimitris Moraitidis Nov 7 '17 at 17:57
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Hint: Try a proof by contradiction. If $f$ were not bounded from above and monotonically increasing, can you show that $\lim_{x\to\infty} f(x)$ does not exist?

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  • $\begingroup$ wouldn't in that case the limit be equal to $\infty$ ? $\endgroup$ – Dimitris Moraitidis Nov 7 '17 at 17:45
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    $\begingroup$ @jimogios Typically, "the limit exists" means "the limit is a finite number." So "the limit is $\infty$" would fall under the category of "the limit does not exist." $\endgroup$ – angryavian Nov 7 '17 at 17:50
  • $\begingroup$ Ok, I got it! Thanks. $\endgroup$ – Dimitris Moraitidis Nov 7 '17 at 17:55

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