Showing that limit exists iff left and right limits are equal Theorem of interest:

"The limit at an interior point of the domain of a function exists if
and only if the left-hand limit and the right-hand limit exist and are
equal 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),
$$
\begin{align}
\left| f\left(x_0 + \frac{\delta}{2}\right) - f\left(x_0+\frac{\delta}{2} \right) \right| & = (x_0+a) - (x_0-a) \cr
\left| f\left(x_0 + \frac{\delta}{2}\right) - L + L - f\left(x_0+\frac{\delta}{2} \right) \right| & = 2a \cr
\left| f\left( x_0 + \frac{\delta}{2} \right) -L \right| + \left| -\left( f\left( x_0 -\frac{\delta}{2}\right)-L \right) \right| &\geq 2a \cr
\frac{1}{2} + \frac{1}{2} &\geq 2a \cr
1 &\ngeq 2a \text{   (contradiction since $a>0$)}
\end{align}
$$
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?
 A: Theorem of interest:

"The limit at an interior point of the domain of a function exists if
and only if the left-hand limit and the right-hand limit exist and are
equal to each other."

Here's a proof, first in the forward, then in the reverse, direction:

*

*Suppose that an accumulation point $c$ of the domain of a function $f$ with
domain $D$ have limit $l,$ i.e.,
$$\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(0{<}|x-c|{<}\delta\implies|f(x)-l|{<}\varepsilon\Big).$$ Then
$$\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(0{<}x-c{<}\delta\:\text{ or
}\:0{<}c-x{<}\delta\implies|f(x)-l|{<}\varepsilon\Big);\\
\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(c{<}x{<}c+\delta\:\text{ or
}\:c-\delta{<}x{<}c\implies|f(x)-l|{<}\varepsilon\Big);\\
\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(\big(c{<}x{<}c+\delta\implies|f(x)-l|{<}\varepsilon\big)\\\text{ and
}\big(c-\delta{<}x{<}c\implies|f(x)-l|{<}\varepsilon\big)\Big);\\
\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(c{<}x{<}c+\delta\implies|f(x)-l|{<}\varepsilon\Big)\\\text{ and
}\,\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(c-\delta{<}x{<}c\implies|f(x)-l|{<}\varepsilon\Big);$$ so, $f$ has
equal left- and right- limits at $c.$

*Suppose that an accumulation point $c$ of the domain of a function $f$ with
domain $D$ have left- and right- limits $l,$ i.e.,
$$\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(c{<}x{<}c+\delta\implies|f(x)-l|{<}\varepsilon\Big)\\\text{ and
}\,\forall\varepsilon{>}0\;\exists\delta{>}0\;\forall x{\in}
D\,\Big(c-\delta{<}x{<}c\implies|f(x)-l|{<}\varepsilon\Big).$$ Then (choosing
$\delta_3=\min(\delta_1,\delta_2)$),
$$\forall\varepsilon{>}0\;\exists\delta_1{>}0\;\exists\delta_2{>}0\;\forall
x{\in}
D\,\Big(\big(c{<}x{<}c+\delta_1\implies|f(x)-l|{<}\varepsilon\big)\\\text{ and
}\big(c-\delta_2{<}x{<}c\implies|f(x)-l|{<}\varepsilon\big)\Big);\\
\forall\varepsilon{>}0\;\exists\delta_1{>}0\;\exists\delta_2{>}0\;\forall
x{\in} D\,\Big(c{<}x{<}c+\delta_1 \:\text{ or
}\: c-\delta_2{<}x{<}c \implies|f(x)-l|{<}\varepsilon\Big);\\
\forall\varepsilon{>}0\;\exists\delta_3{>}0\;\forall x{\in}
D\,\Big(0{<}|x-c|{<}\delta_3\implies|f(x)-l|{<}\varepsilon\Big);$$ so, $f$
has a limit at $c.$
