Show $f(x)$ is bounded in a neighbourhood of its limit points The attempt I made doesn't cover the case for $x=c$. How can I make it so it does?
Prove that if a function $f : A \to \mathbb{R} $ has a limit $l \in \mathbb{R} $ at $c \in L(A)$, then it is bounded in a neighborhood of $c$, i.e there exists $M \in \mathbb{R}$ and $\delta > 0$ such that for any $x \in (c-\delta,c+\delta) \cap A$, $\|f(x)\| \leq M$
Note: $L(A)$ signifies the set of limit points in $A$.
And these results use end points in definitions for limits.
My attempt:
Since $f$ has a limit $l \in \mathbb{R}$ at a point $c \in L(A)$, then by definition $\forall \epsilon >0 \space \exists \space \delta \space \text{s.t if} \space x \in A \space \text{and} \space 0 < |x-c| \leq \delta \space \text{it holds that} |f(x)-l| \leq \epsilon$
Then $$l - \epsilon \leq f(x) \leq \epsilon + l$$
if $$c - \delta \leq x \leq \delta + c$$
Thus if $$M = max(l - \epsilon, l+ \epsilon)$$
then when $$x \in (c-\delta, c+\delta) \cap A$$
$$|f(x)| \leq M$$
Is this a sufficient/correct proof?

To put it in a better way let us choose $\epsilon = 1$. Then corresponding to this $\epsilon$ you will get a $\delta>0$ such that $|f(x)-l|<\epsilon$ for all $x\in (c-\delta,c+\delta)\cap A$ and $x\neq c$ (just the definition of limit). In particular, you have
\begin{equation}
|f(x)| = |f(x) -l+l|\leq |f(x)-l|+|l| <1+|l|,
\end{equation}
for all $x\in (c-\delta,c+\delta)\cap A$, $x\neq c$. 
So, if $c\notin (c-\delta,c+\delta)\cap A$, then choose $M = |l|+1$ and $\delta$ as above to get $|f(x)|\leq M$ for all $x\in (c-\delta,c+\delta)\cap A$.
If $c \in (c-\delta,c+\delta)\cap A$, then choose $M = \max\{|f(c)|,|l|+1\}$ and $\delta$ as above to get $|f(x)|\leq M$ for all $x\in (c-\delta,c+\delta)\cap A$.
