Theorem of interest:
The limit at an interior point of the domain of a function exists ifand only if the left-hand limit and the right-hand limit exist and areequal to each other.
I'm using the $ \epsilon$-$\delta $ definition of limits and will be proving by contradiction.
Let $f$ be a function of $x$ defined within an open interval around $x_0$ (where $x_0$ need not be defined). Let the limit of $f$ as $x$ approaches $x_0$ be $L$. Let $f$ be:$$ f(x) = \cases{ x_0 +a & $x > x_0$ \cr x_0 -a & $x<x_0$}$$where $a>0 , a \in\Re $.
Suppose the limit $L$ does exist. Let $\epsilon = \frac{1}{2}$ (an arbitrary small number) and $\delta = \delta_{\epsilon}>0$.
Since $L$ exists then it must be that for all $x \in \left( x_0-\delta, x_0+\delta\right)$, we have $\left| f(x) - L\right|<\epsilon = \frac{1}{2}$. However, at $x=x_0+\frac{\delta}{2}$ and $x=x_0-\frac{\delta}{2}$ (random values within given domain of f),$$\begin{align}\left| f\left(x_0 + \frac{\delta}{2}\right) - f\left(x_0+\frac{\delta}{2} \right) \right| & = (x_0+a) - (x_0-a) \cr\left| f\left(x_0 + \frac{\delta}{2}\right) - L + L - f\left(x_0+\frac{\delta}{2} \right) \right| & = 2a \cr\left| f\left( x_0 + \frac{\delta}{2} \right) -L \right| + \left| -\left( f\left( x_0 -\frac{\delta}{2}\right)-L \right) \right| &\geq 2a \cr\frac{1}{2} + \frac{1}{2} &\geq 2a \cr1 &\ngeq 2a \text{ (contradiction since $a>0$)}\end{align}$$Hence, we can see that because of the fact that the limit L does not exist, the arithmetic between the functions resulted in a contradiction. However, I'm not sure if the said proof is rigorous (not sure what that means). I picked $\epsilon=\frac{1}{2}$ because in limits we are playing on the notion of $x$ approaching a certain value, hence epsilon is logically small. If epsilon is large enough there wouldn't be (or at least, inconclusive) a contradiction as we see above.
So, did I miss anything?