It is first necessary to distinguish between interior and boundary points of the domain of a function $f$.

A point P is an interior point of the domain of a function $f$ if P is contained in a neighborhood that lies completely in the domain of $f$.

A point P is a boundary point of the domain of a function $f$ if every neighborhood of P contains points that are in the domain and points that are not in the domain.

Definition 3.2.1 formalizes the meaning of a limit in the plane, called in this guide, the bivariate limit.

The points satisfying $0<\sqrt{{\left(x-a\right)}^2\&plus;{\left(y-b\right)}^2}<\mathrm{\delta }$ are said to lie in a deleted neighborhood of $\left(a\&comma;b\right)$. This deleted neighborhood is actually the interior of an annulus that is an open disk of radius $\mathrm{\&delta;}$ with the center point $\left(a\&comma;b\right)$ removed.

Conceptually, Definition 3.2.1 is the generalization of the formal definition of a limit along the real line: the values of $f$ can be made arbitrarily close to $L$ by the expedient of taking $\left(x\&comma;y\right)$ sufficiently close to P. Thus, $L$ is the limit of $f\left(x\&comma;y\right)$ at P if all the points in the deleted neighborhood of P produce function values that are close to $L$.

The distinction between interior and boundary points amounts to this: The limit is taken over the domain of the function.

It is generally difficult to prove that $L$ is the limiting value of $f\left(x\&comma;y\right)$ because the requisite estimates demand a facility with manipulating inequalities.

Table 3.2.1 lists several inequalities that are useful for proving that a real number $L$ is indeed the bivariate limit of $f\left(x\&comma;y\right)$. Inequality 3 is the "triangle" inequality. Inequalities 4, 5, 6, and 7 should be self-evident. A proof of Inequality 2 uses Inequality 1, which itself is proved in Example 3.2.28.

It is far easier to show that $f\left(x\&comma;y\right)$ does not have a limit at $\left(a\&comma;b\right)$. Recall that for a limit on the line to exist, both left-hand and right-hand limits must exist, and be equal. The same idea holds for the bivariate limit. If the limits taken along two different paths are not equal, then the (bivariate) limit cannot exist. Consequently, to show a bivariate limit does not exist at $\left(a\&comma;b\right)$, it suffices to show that along two different paths through $\left(a\&comma;b\right)$ the limits differ, or that one such limit does not exist.

Typical trial paths are the axes, lines $y\&equals;m x$, parabolas $y\&equals;x^2$ and $x\&equals;y^2$, and on rare occasions, the curves $y\&equals;x^{3\&sol;2}$ and $y\&equals;x^{2\&sol;3}$.