Why is this function not unbounded? 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?
 A: 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.)
A: 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.
