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Suppose we have a real valued function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$ \forall \epsilon > 0, \exists \delta \ s.t. \ |x-1| \ge \delta \rightarrow |f(x) - f(1)| \ge \epsilon $$

Then it is claimed in my exam that

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

BUT

$$ f \ \text{is not unbounded}$$

I think these statements are contradictory, as $f$ is bounded if $\exists M \in \mathbb{R}, \text{ s. t. } \forall u \in \mathbb{R} \ |f(u)| < M $

But if: $$ \lim_{|x| \rightarrow \infty} |f(x)| = \infty$$

Then $\exists$ a sequence of values $u_1 , u_2 ... $ such that $f(u_i)$ is an unbounded sequence.

What am I missing here?

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  • $\begingroup$ I posted an answer arguing why $f$ should be unbounded without fully reading your question; sorry about that. You're absolutely right, if $f$ were bounded we can't have $\lim \limits_{|x| \to \infty} |f(x)| = \infty$. $\endgroup$ – layman Sep 10 '16 at 21:45
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Well, I can see why $\lim_{|x| \rightarrow \infty} |f(x)| = \infty $:

In $\forall \epsilon > 0, \exists \delta \ s.t. \ |x-1| \ge \delta \rightarrow |f(x) - f(1)| \ge \epsilon $, choose $\epsilon$ very large and you get $|f(x) - f(1)| \ge \epsilon$ so $|f(x)| \ge \epsilon- |f(1)| $.

I agree with you that "$f \ \text{is not unbounded} $", which, to me, is the same as $f$ is bounded, contradicts this.

So, I am not sure what is going on here.

Go ahead, somebody: Prove me wrong. (Won't be the first time.)

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  • $\begingroup$ I don't think you're wrong. Either OP misinterpreted the exam question, or there is a typo on the exam. $\endgroup$ – layman Sep 10 '16 at 21:46
  • $\begingroup$ ets.org/s/gre/pdf/practice_book_math.pdf, Question 60, page 54. $\endgroup$ – frogeyedpeas Sep 10 '16 at 21:52
  • $\begingroup$ Actually I think i know the issue, "equivalent" is the condition they are harping on. $\endgroup$ – frogeyedpeas Sep 10 '16 at 21:54
  • $\begingroup$ @frogeyedpeas Where in that question do they suggest $f$ is bounded? $\endgroup$ – layman Sep 10 '16 at 21:55
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    $\begingroup$ @frogeyedpeas Hmm, are you sure? I was thinking more along the lines of something that is unbounded as you approach $x=1$. Like taking $f(x) = \begin{cases} \tan(\frac{\pi x}{2}) & 0 \leq x < 1 \\ 3 & x \not \in [0,1) \end{cases}$, or something like that. That way, the only way you approach infinity is as $x$ approaches $1$ (from the left in this case). $\endgroup$ – layman Sep 10 '16 at 22:07
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First, note that $f(1)$ is a constant.

By the definition of your function, we are assured no matter how big we make $\epsilon$, we can find some $\delta$ such that if $x$ is more than $\delta$ away from the value $1$, then $f(x)$ will be more than $\epsilon$ away from $f(1)$.

So, we can make $\epsilon$ as large as we want, and we are assured we can still find $x$ values such that $f(x)$ is at least $\epsilon$-distance from $f(1)$. I hope it's clear from this that $f$ must then be unbounded.

If it were bounded, then we know for all $x$, $f(x)$ must be in some set $[-M,M]$ for some positive $M$. But if that's true, then $f(x)$ cannot be more than $2M + 1$ away from $f(1)$ (why?). But we just said we can always find $x$ values so that $f(x)$ is as far away as we want from $f(1)$. So that's a contradiction, and so $f$ is unbounded.

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