The Problem:
Show that if $\lim_{x \to a}f(x)=L$ then $f$ is bounded near $a$, i.e. there are constants $C,M > 0$ such that $\left|f(x)\right|<M$ for all $x$ such that $\left|x-a\right| < C$.
Where I Am:
I feel like this question is more simple than I'm making it out to be, but it just seems to be worded in a confusing way. I assume that it's a question about continuity, and therefore the answer simply follows from the definition of continuity, namely, that $f$ is continuous at a point $a$ if
$$ \forall \epsilon, \exists \delta \text{ such that } \left| x-a \right| < \delta \implies \left| f(x)-f(a) \right| < \epsilon.$$
So, in this case $C$ and $M$ are kind of like $\delta$ and $\epsilon$, respectively, where, I guess, $f(a) = 0$. Furthermore, there's a neighborhood, i.e. an open interval, $(a-C,a+C)$ in which $L$ lies, i.e. $L$ is "bounded" by this interval, to use the wording of the question. This, however, is clearly not a proof, which I need to provide. So, if someone could offer a little guidance here, it'd be greatly appreciated. Thanks!
Let $\epsilon =1$ so there is $\delta>0$ such that $|x-a|<\delta\implies |f(x)-f(a)|<1$ and then $|f(x)|<|f(a)|+1=M$. Hence for $x\in(a-\delta,a+\delta)$ we have $|f(x)|\le M$ and the result follows.