Continuity in Functions of Two Variables f(x,y)

A function of two or more variables \(f: \mathbb{R}^n \to \mathbb{R}\) is said to be continuous at a point \(\vec{x}_0 \in \mathbb{R}^n\) if the limit of the function at \(\vec{x}_0\) coincides with the value of the function at that point: \[ \lim_{ \vec{x} \rightarrow \vec{x}_0 } f( \vec{x} ) = f( \vec{x}_0 ) \] where \(\vec{x}_0\) is a vector consisting of \(n\) real components.
 

For example, a function of two variables f(x, y) is continuous at (x₀, y₀) if the limit as (x, y) approaches (x₀, y₀) equals f(x₀, y₀).

$$ \lim_{(x,y) \rightarrow (x_0,y_0)} f(x,y) = f(x_0,y_0) $$

In this case, \((x_0, y_0)\) is simply an ordered pair of real numbers, rather than a vector in the formal sense.

However, we can treat the pair (x₀, y₀) as a two-dimensional vector:

$$ \vec{x}_0 = \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} $$

Likewise, any generic pair (x, y) can be represented as a two-component vector:

$$ \vec{x} = \begin{pmatrix} x \\ y \end{pmatrix} $$

By expressing the limit in terms of vectors, we obtain what’s known as the vector form of the limit:

$$ \lim_{\vec{x} \rightarrow \vec{x}_0} f( \vec{x} ) = f( \vec{x}_0) $$

This definition is entirely equivalent to the previous one expressed in terms of scalar pairs.

Note. The vector-based definition of a limit closely mirrors the standard limit definition for a single-variable function f(x). The key difference is that here, the input \( \vec{x} \) is a vector with two components.

Why prefer the vector definition?

The vector notation generalizes naturally to functions of three variables, such as f(x, y, z), and more generally to functions of n variables, like f(x₁, x₂, ..., x_n). This makes the notation both more concise and more powerful in higher dimensions.

A Practical Example

Consider the limit of the function f: ℝ² → ℝ given by

$$ \lim_{(x,y) \rightarrow (0,0) } \sin y + \cos xy $$

We can evaluate this by separating the limit into two parts:

$$ \lim_{(x,y) \rightarrow (0,0) } \sin y + \lim_{(x,y) \rightarrow (0,0) } \cos xy $$

As (x, y) approaches (0, 0), we observe that sin y tends to 0, while cos(xy) approaches 1.

$$ \lim_{(x,y) \rightarrow (0,0) } \sin y + \lim_{(x,y) \rightarrow (0,0) } \cos xy = 0 + 1 = 1 $$

Therefore, the limit of the function f(x, y) exists and is equal to 1:

$$ \lim_{(x,y) \rightarrow (0,0) } \sin y + \cos xy = 1 $$

Substituting x = 0 and y = 0 directly into the function yields the same result:

$$ f(0,0) = \sin y + \cos xy = \sin 0 + \cos 0 = 0 + 1 = 1 $$

Since the limit and the function value agree, we conclude that the function is continuous at the point (0,0).

$$ \lim_{(x,y) \rightarrow (0,0) } \sin y + \cos xy = f(0,0) = 1 $$

From a geometric perspective, \(f(x, y)\) defines a surface in three-dimensional space \((x, y, z)\), where \(z = f(x, y)\). 

Continuity at the point \((x_0, y_0)\) is reflected in the fact that the surface is smooth at the corresponding point \((x_0, y_0, f(x_0, y_0))\), with no gaps, jumps, or sharp corners.

And so on.