Question: Suppose that $f : \mathbb{R} →\mathbb{R}$ and $\lim\_{x\to a} f(x) = L$ (L-finite). Show that $f$ must be bounded in an open interval centered at $x = a$.

---

By definition of function limits.

$\forall\epsilon >0$, there $\exists \delta \in \mathbb{R}$ s.t if $x\in D$ and $0<|x-a|< \delta$, then $|f(x)-L|<\epsilon$

consider the open interval $(a-\delta ,a+\delta)$ thus by above, f is bounded above by $max(L+\epsilon,f(a))$ and bounded below by $min(L-\epsilon,f(a))$ Q.E.D

---

My professor gave me some tips of how to get the end goal however there is still some strings I need to tie. I'm getting somewhat twisted when it says $x=a$. Does the open interval $(a-\delta ,a+\delta)$ have anything to do with the definition that is "$0<|x-a|< \delta$" or is the open interval separate from the definition at this point? and should I consider a variable change?

Answer: 
The open interval is precisely the values of $x$ for which $|x-a|<\delta$; if you restrict it to positive absolute values, you punch a hole in the open interval at $a$. Since the function has a definite value at $a$, however, if you've shown that the function is bounded for the punctured interval, you can extend the bound to encompass $f(a)$.
If $|f(x)| \leq M$ for $0<|x-a|<\delta$ then $|f(x)| \leq C$ for all $x \in (a-\delta, a+\delta)$ where $C$ is the maximum of $M$ and $|f(a)|$.