Continuity of Multivariable Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.2: “Limits and Continuity” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-2-limits-and-continuity |
| Textbook used in class | Stewart, Calculus, Section 14.2: “Limits and Continuity” (Examples 4, 5) |
Quick Reference
Definition: $f$ is continuous at $(a,b)$ if:
- $f(a,b)$ is defined,
- $\displaystyle\lim_{(x,y)\to(a,b)} f(x,y)$ exists, and
- $\displaystyle\lim_{(x,y)\to(a,b)} f(x,y) = f(a,b)$.
Polynomial functions are continuous everywhere.
Rational functions are continuous wherever the denominator is nonzero.
Compositions of continuous functions are continuous.
Motivation
Continuity for functions of two variables carries the same intuition as in one variable: a function is continuous if small changes in the input produce small changes in the output, with no jumps or breaks. The definition is identical in spirit -- the limit as $(x,y)$ approaches $(a,b)$ must equal the function value there.
The practical payoff is that checking continuity for most common functions is easy: polynomials and compositions of elementary functions (exponentials, trigonometric, etc.) are continuous wherever they are defined. Only at potential points of discontinuity (like where a denominator vanishes) do you need to check carefully.
Key Concepts
1. Applying the Definition
For most functions encountered in calculus, continuous functions composed with continuous functions remain continuous. So $f(x,y) = e^{x^2+y^2}$, $g(x,y) = \sin(xy)$, and $h(x,y) = x^2 y + y^3$ are all continuous everywhere.
The only concern arises at points where the formula breaks down (division by zero, square root of a negative, etc.).
2. Piecewise-Defined Functions
When a function is defined by different formulas on different regions, you must check continuity at the boundary. The typical approach: compute the limit from the previous skill and compare to the function value at the boundary point.
Worked Example
Determine where $f(x,y) = \dfrac{x^2 - y^2}{x^2 + y^2}$ is continuous. (Stewart 14.2.)
The numerator and denominator are polynomials, hence continuous everywhere. The denominator $x^2 + y^2 = 0$ only at $(0,0)$. At all other points, $f$ is a ratio of continuous functions with a nonzero denominator, so $f$ is continuous on $\mathbb{R}^2 \setminus \{(0,0)\}$.
At $(0,0)$: we showed in the limits lesson that the limit does not exist, so $f$ cannot be made continuous at the origin.
Example 2. Is $g(x,y) = \begin{cases} \dfrac{x^2 y}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$ continuous at $(0,0)$?
From the limits lesson: $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^2+y^2} = 0 = g(0,0)$.
All three conditions of the definition hold, so $g$ is continuous at the origin.
if partial derivatives exist at a point, the function is continuous there. Partial derivatives can exist without the function being continuous. Consider $f(x,y) = \dfrac{xy}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$. The partial derivatives $f_x(0,0)$ and $f_y(0,0)$ both equal 0 (check by definition), yet the limit of $f$ at $(0,0)$ does not exist, so $f$ is not continuous there. Continuity requires the multivariable limit, not just the along-axis behavior.
Leveled Practice
Problem 1. Where is $f(x,y) = \dfrac{\sin(xy)}{x^2 + y^2 + 1}$ continuous?
Show answer
$\sin(xy)$ is continuous everywhere. $x^2+y^2+1 \geq 1 > 0$ everywhere (denominator never zero). So $f$ is continuous on all of $\mathbb{R}^2$.