# Proving a limit does not exist with epsilon delta I don't have an equation for this. I'm looking more for intuition rather than anything else. At what point in the epsilon-delta definition of a limit does this fail to hold up?

I'm having trouble understanding the epsilon-delta definition of a limit and why it's defined the way it is.

Call the size of the jump $a$. Choose $\varepsilon=a/2$. Then no matter how small you choose $\delta$, there will be $x$ (on the other side of the jump from $x_0$) such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|>\varepsilon$.
I'll call your function $f$ for simplicity.
It would help if the $y$-axis were labeled. Write $y_1$ for the open dot (which is not equal to $f(x_0)$) and $y_2 = f(x_0)$ for the filled-in dot. Then $y_1 - y_0$ is some nonzero number; since you only care about the idea, I'll just say that number is $2$.
Let $\epsilon = .5$. If the function were continuous at $x_0$, that would say that there exists $\delta$ such that, for any $x$ within $\delta$ of $x_0$, $f(x)$ is within $.5$ of $f(x_0) = y_2$.
But that's not true. Given your $\delta$, I take a positive number $y$ that's less than $x_0$, but closer to $x_0$ than $x_0 - \delta$. Then $y$ is within $\delta$ of $x_0$. From the picture, $f(y) < y_1$. But now clearly $f(y)$ is not within $.5$ of $y_2$.
Since no $\delta$ exists, the function can't be continuous. (Easy exercise: modify this proof where we drop the assumption that the distance is $2$ but just call it $y_1 - y_0$.)