Using circle rather than rectangle to define limit precisely Take a look at the following precise definition for limit

The geometric representation is 

It is lucid that the definition of limit is the ability to find such rectangle (i.e. the intersect of blue and orange rectangles), if so, then the limit does exist. My question is is it possible to formulate the definition using circle rather than rectangle and if so how? 

 A: First and foremost, no this isn't really possible, due to the fact that the "rectangular" definition most definitely needs to be a rectangle. You cannot force the definition to be "square", because this would enforce us to always choose $\delta \ge \varepsilon$. For many functions, (e.g. $f(x) = 2x$), this will not be appropriate, and you'd obtain a strictly weaker form of limit (one where $f(x) = 2x$ would have no limits as $x$ approached any point).
A "circle" shape has much the same problem: it doesn't allow for $\delta$ to be smaller than $\varepsilon$ (or whatever the "circular" counterparts would be), which it must be allowed to do.
However, I think we could do an elliptical version. That would be the obvious generalisation of the circle that we could work with.
Essentially, we want to start with a $\forall \varepsilon > 0$ as per usual, but we need to limit our function values differently. We need to not simply have $|f(x) - L|$ be less than $\varepsilon$, but have it be less than a function of $\varepsilon$ and $x$, or more precisely, the distance from $x$ to the point its approaching (let's say $a$). That function needs to draw out an ellipse with one axis width $\delta$, and the other width $\varepsilon$. Here's what we get:

For all $\varepsilon > 0$, there exists some $\delta > 0$ such that
  $$0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon\sqrt{1 - \frac{|x - a|^2}{\delta^2}}.$$

So, that's the "how". The above should be equivalent to the "rectangular" definition. I'll leave the "why" (or more particularly, "why not") as an exercise.
