Answer
$$f(x)=\begin{cases}x, & x \ge 0 \\ x+1 & x<0\end{cases} \qquad g(x)=\begin{cases}-x, & x \ge 0 \\ -x-1 & x<0\end{cases}$$
Work Step by Step
Consider the following functions:$$f(x)=\begin{cases}x, & x \ge 0 \\ x+1 & x<0\end{cases} \qquad g(x)=\begin{cases}-x, & x \ge 0 \\ -x-1 & x<0\end{cases}.$$It is clear that both functions do not have a limit at $x=0$ since they have a jump there; in other words, the two functions do not approach a specific number when $x$ comes close to $0$.
However, the sum of the functions does have a limit at $x=0$, so the resultant function is a constant function, which has a limit at every point, that is,$$f(x)+g(x)=0 \quad \Rightarrow \quad \lim_{x \to 0}[f(x)+g(x)]=0.$$
