DFA that accepts two more $1$'s than $0$'s
As mentioned this is not a regular language.

To prove this, recall that a language, $L$, is regular iff some DFA recognizes it. 

We will show that no DFA recognizes $L$.
Consider the infinite set of strings $S=\{0^n : n \ge 0\}$.
Then any two strings in $S$ are of the form $\alpha=0^x$ and $\beta=0^y$, $x,y\ge0$, we may assume wlog that $x&gt;y$.

Let $\lambda=1^{y+2}$.

Then $\alpha\lambda=0^x1^{y+2} \notin L$ (since $x&gt;y \implies y+2 -x &lt; 2$), but $\beta\lambda=0^y1^{y+2} \in L$.

Thus we have an infinite set of strings (prefixes), $S$, such that for every pair of distinct strings $\alpha,\beta \in S$ there exists another string (suffix) $\lambda$ where exactly one of $\alpha\lambda$ or $\beta\lambda$ is in $L$. Thinking about this, this means that if a DFA, $D$, recognizes $L$, then every string in $S$ must take $D$ to a different state (if $\alpha,\beta \in S$ both took $D$ to the same state then so would $\alpha\lambda$ and $\beta\lambda$ for any $\lambda$, but we know that only one of these strings is in $L$). So $D$ must have at least $|S|$ states, but $|S|=\infty$, thus $D$ cannot exist.