Is there any way for a function to be unbounded in a neighborhood of $a$ besides division by $0$? The standard examples of function that are unbounded in a neighborhood of $a$ are functions of the $\frac{1}{x-a}$ variety, or $\tan(x)$, which is just division by $0$ but hidden a bit better.
Are there other examples of functions that are unbounded in a neighborhood of a point without any division by 0?
 A: $\ln\lvert x\rvert$ for $a=0$ comes to mind.
A: Since you didn't say continuous, here is one example:
$$f(x)=\begin{cases}0\text{ if }x\text{ is irrational}\\q\text{ if }x=\frac{p}{q}\text{ in lowest terms}\end{cases}$$
For a continuous, piecewise defined function, consider linearly interpolating between the points $(\tfrac{1}{n},n)$ for $n=1,2,\dots$
A: For the sake of simplicity suppose that $\lim_{x\to0}f(x)=+\infty$, $f(x)\ne0$ if $x\ne0$. Let
$$
g(x)=\begin{cases}1/f(x) & x\ne0,\\0 &x=0.\end{cases}
$$
Then
$$
f(x)=\frac{1}{g(x)},\quad x\ne0.
$$
We see that we can always with any infinite limit as division by $0$.
A: One way to build a fairly wild unbounded function is through measure theory. Let $f(t) = t^{-1/2}\cdot \chi_{(0,1)}.$ Let $q_1,q_2, \dots$ be the rationals. Then the function
$$g(t) = \sum_{n=1}^{\infty} 2^{-n}f(t-q_n)$$
is is finite a.e., and because $g$ blows up at each $q_n,$ $g$ is unbounded on every nonempty open subinterval of $\mathbb R.$
Here is a more ambitious example. Claim: There exists a function $f:\mathbb R\to \mathbb R $ such that $f(I) =\mathbb R$ for every nonempty open subinterval $I$ of $\mathbb R.$
Proof: Let $I_1,I_2, \dots$ be the open subintervals of $\mathbb R$ with rational end points. We can inductively choose pairwise disjoint Cantor sets $K_1,K_2,\dots$ such that $K_n\subset I_n$ for each $n.$
Now each $K_n$ has the cardinality of $\mathbb R.$ Hence for each $n$ there is a bijection $f_n: K_n \to \mathbb R.$ Define $f:\mathbb R \to \mathbb R$ by setting $f= f_n$ on each $K_n,$ $f=0$ everywhere else. (Since the $K_n$ are pairwise disjoint, this map is well defined.)
Let $I$ be a nonempty open subinterval of $\mathbb R.$ Then $I$ contains some $I_n.$ Hence it contains $K_n.$ Thus $f((a,b))$ contains $f(K_n) =  f_n(K_n) = \mathbb R.$ Thus the claim is proved.
