If the limit of a real-valued function on $[a,b]$ exists on every point of the interval, is it necessary to be bounded on $[a,b]$? Let $f$ be a real-valued function on $[a,b]$, and the limit of $f$ exists on every point of $[a,b]$, then is $f$ bounded on this interval?

I guess $f$ is not necessary to be bounded but I couldn't find out an counterexample. 
 A: Yes, it must be bounded.
Suppose that $f\colon[a,b]\longrightarrow\mathbb R$ is unbounded. For each $n\in\mathbb N$, let $x_n\in[a,b]$ be such that $f(x_n) \geqslant n$. By the Bolzano-Weierstrass theorem, the sequence $(x_n)_{n\in\mathbb N}$ has a convergent subsequence and we can assume without loss of generality that the whole sequence $(x_n)_{n\in\mathbb N}$ converges to some $x_0\in[a,b]$. But then $\lim_{n\to\infty}f(x_n)$  does not exist (in $\mathbb R$) and therefore the limit $\lim_{x\to x_0}f(x)$ does not exist.
A: Correct me if wrong.
Let me rephrase this a bit:
$A: = \sup${$f(x); x \in [a,b]$ } $\in R \cup {\infty}.$
$(A = \infty$ , if $f$ is not bounded above).
1) If $A \in \mathbb{R}$ we are done.
2) Assume $f$ is not bounded on $[a,b]$.
There is a sequence $y_n \in f([a,b])$ such that for every $M$, positive, real, there is a $n_0 \in \mathbb{N}$ such that for  $n \ge n_0$, $ n \in \mathbb{N}$, we have  $y_n \gt M.$
Let $x_n$ be such that $y_n = f(x_n).$
Since the sequence $x_n$ is bounded, $x_n \in [a,b]$, by Bolzano Weierstrass there is a convergent subsequence
$x_{n_k}$ with 
$\lim_{k \rightarrow \infty} x_{n_k} = p \in [a,b],$ since [a,b] compact.
By assumption : 
$ \lim_{ k \in \mathbb{N}} f(x_{n_k}) = L$ $\in \mathbb{R}$ exists.
Contradiction.
Hence $f$ is bounded.
