I think a good example is the following function which Spivak presents in his chapter devoted to Limits.
Define $f:(0,1)\to\mathbb R$ as $f(x)=\begin{cases} 0\text{ ; if }x \text{ is irrational}\\\frac 1 p \text{ ; if } x=\frac q p \text{ an irreducible fraction}\end{cases}$
This function is very "strange". You can try plotting it for some values on $[0,1]$. Try taking $q=1,2,3,\dots$ to make it easier. The functions has a nice pattern.
What we want to prove is that, for every $a\in(0,1)$, we have $\lim_{x\to a}f(x)=0$
from where we'll see it is only continuous at the irrational points. So, let's pick any $a\in(0,1)$, and let us be given $\epsilon >0$. Choose $n$ so that $1/n\leq \epsilon$.
First, we note that the only points where it might be false that $|f(x)-0|<\epsilon$ are $\frac{1}{2};\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}; \cdots ;\frac{1}{n}, \cdots ,\frac{{n - 1}}{n}$ If $a$ is rational, then $a$ could be one of those numbers. But, as many as there can be, the amount of these numbers are finite. Thus, for some $p/q$ in that list, the number $|a-p/q|$ is least. If $a$ is one of these numbers, consider $p/q\neq a$. Then, take $\delta$ as this distance. It will then be the case that if $0<|x-a|<\delta$, then $x$ will be none of the numbers
$\frac{1}{2};\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}; \cdots ;\frac{1}{n}, \cdots ,\frac{{n - 1}}{n}$
thus it will be true that $|f(x)-0|<\epsilon$.
Another example, closely related to this one, is:
For each $n$, let $A_n$ be a finite subset of $(0,1)$. That is, each $A_n$ is a set consisting of numbers with $0, and let them be such that if $n\neq m$, $A_n$ and $A_m$ have no elements in common. We define then $f$ as
$f(x)=\begin{cases} 0\text{ ; if }x \text{ is not in any }A_n \\\frac 1 n \text{ ; if } x\in A_n\end{cases}$
Now, we want to prove again that, for every $a\in(0,1)$, we have $\lim_{x\to a}f(x)=0$
Now, if $\lim\limits_{x\to a}f(x)=0$, this means that for every $\epsilon >0$ there is a $\delta >0$ such that $0<|x-a|<\delta$ implies $|f(x)|<\epsilon$. Since $A_1$ is finite, there must exist a $\delta_1>0$ such that no $x\in A_1$ is in $(a-\delta_1,a+\delta_1)-\{a\}$. Thus, we must have $|f(x)|\leq 1/2$ for any $x$ with $0<|x-a|<\delta_1$. Similarily there exists a $\delta_2>0$ such that no $x\in A_2$ is in $(a-\delta_2,a+\delta_2)-\{a\}$ from which it is $ |f(x)|\leq 1/3$ in $0<|x-a|<\delta_2$. Analogously, there exists a $\delta_n >0$ such that no $x\in A_n$ is in $(a-\delta_n,a+\delta_n)-\{a\}$ from where $|f(x)|\leq 1/n$ for $0<|x-a|<\delta_n$. We're ready to finish it off. Let $\epsilon >0$ be given and choose $n_0$ such that $1/n_0\leq \epsilon$. Let $n\geq n_0$ and $\delta =\min\{\delta_1,\dots,\delta_n\}$. Then, whenever $0<|x-a|<\delta$, we must have $|f(x)|<\epsilon$, from where it must be $\lim_{x\to a}f(x)=0$ for every $a\in(0,1)$.
Note how in this two examples, the finiteness of the sets in the second case and of the "bad" numbers in the first was key. Note also that we really have no explicity formula for $f$ that allows any algebraic manipulation whatsoever. (We can actually put $f\left( x \right) = \sum\limits_{n \in {\Bbb N}} {\frac{1}{n}{\chi _{{A_n}}}\left( x \right)} $ in the second case, but not much is revealed).
Let alone is the recommendation that you read Spivaks exposition, if you can.