Math Core

In one variable, $x$ can approach $a$ from only two directions: the left and the right. In the plane, a point $(x, y)$ can approach $(a, b)$ along infinitely many lines and curves. That extra freedom makes limits of functions of two variables more delicate, and it is the main thing to understand before defining derivatives in several variables.

What the limit means

Definition

Limit of a function of two variables

We write $\displaystyle\lim_{(x, y) \to (a, b)} f(x, y) = L$ if the values $f(x, y)$ get arbitrarily close to $L$ whenever $(x, y)$ is close enough to $(a, b)$ , with $(x, y) \ne (a, b)$ . Precisely: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ implies $\lvert f(x, y) - L\rvert < \varepsilon$ .

The key phrase is "whenever $(x, y)$ is close enough." The definition does not care which route $(x, y)$ takes. Every route into $(a, b)$ has to produce the same limit $L$ .

Continuous functions: substitute

A function is continuous at $(a, b)$ if $\displaystyle\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b)$ . The limit laws you know (sums, products, quotients with nonzero denominators, compositions) all hold in two variables, so the following functions are continuous wherever they are defined:

polynomials in $x$ and $y$ , such as $x^3 y - 5xy^2 + 7$ ,
rational functions (quotients of polynomials) wherever the denominator is nonzero,
compositions of continuous functions, such as $e^{xy}$ , $\sin(x^2 + y)$ and $\ln(1 + x^2 + y^2)$ .

For these, the limit is found by direct substitution.

Worked example: Direct substitution

Evaluate $\displaystyle\lim_{(x, y) \to (1, 2)} \frac{x^2 y + 3}{x + y}$ .

This is a rational function, and the denominator at $(1, 2)$ is $3 \ne 0$ . Substitute:

\frac{(1)^2(2) + 3}{1 + 2} = \frac{5}{3}.

When substitution gives $\tfrac{0}{0}$ , algebra sometimes helps, just as in one variable. For example,

\lim_{(x, y) \to (2, 2)} \frac{x^2 - y^2}{x - y} = \lim_{(x, y) \to (2, 2)} (x + y) = 4,

because the factor $x - y$ cancels at every point off the line $y = x$ . The points on that line are outside the domain of the original function, and a limit only considers points in the domain, so the cancellation is valid.

Showing a limit does not exist: two paths

If $f(x, y) \to L_1$ along one path into $(a, b)$ and $f(x, y) \to L_2$ along another, with $L_1 \ne L_2$ , then the limit does not exist.

The two-path test

To show $\displaystyle\lim_{(x, y) \to (a, b)} f(x, y)$ does not exist, find two paths into $(a, b)$ that give different limiting values. Good paths to try at the origin: the $x$ -axis ( $y = 0$ ), the $y$ -axis ( $x = 0$ ), lines $y = mx$ , and parabolas $y = x^2$ or $x = y^2$ .

Worked example: Different limits along the axes

Show that $\displaystyle\lim_{(x, y) \to (0, 0)} \frac{x^2 - y^2}{x^2 + y^2}$ does not exist.

Along the $x$ -axis, $y = 0$ and $x \ne 0$ : the function equals $\dfrac{x^2}{x^2} = 1$ .

Along the $y$ -axis, $x = 0$ and $y \ne 0$ : the function equals $\dfrac{-y^2}{y^2} = -1$ .

Two paths give $1$ and $-1$ , so the limit does not exist.

Worked example: Lines are not enough

Investigate $\displaystyle\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^4}$ .

Along any line $y = mx$ (with $x \ne 0$ ):

\frac{x(m^2x^2)}{x^2 + m^4x^4} = \frac{m^2 x}{1 + m^4 x^2} \to 0.

Along the $y$ -axis the function is $0$ as well. It is tempting to conclude the limit is $0$ , but try the parabola $x = y^2$ (with $y \ne 0$ ):

\frac{y^2 \cdot y^2}{y^4 + y^4} = \frac{y^4}{2y^4} = \frac{1}{2}.

Along this parabola the value is always $\tfrac{1}{2}$ . Since $0 \ne \tfrac{1}{2}$ , the limit does not exist.

Common mistake

Getting the same value along many paths never proves that a limit exists. There are infinitely many paths, and the example above shows a function that agrees along every line yet fails along a parabola. Paths can only prove that a limit does not exist. To prove existence, use continuity, algebra, or a squeeze.

Showing a limit exists: squeeze and polar coordinates

If $\lvert f(x, y) - L\rvert \le g(x, y)$ and $g(x, y) \to 0$ , then $f(x, y) \to L$ . This is the Squeeze Theorem in two variables. The trick is usually to bound a fraction by noticing that $x^2 \le x^2 + y^2$ , so $\dfrac{x^2}{x^2 + y^2} \le 1$ .

Worked example: A squeeze

Evaluate $\displaystyle\lim_{(x, y) \to (0, 0)} \frac{3x^2 y}{x^2 + y^2}$ .

For $(x, y) \ne (0, 0)$ ,

\left\lvert \frac{3x^2 y}{x^2 + y^2} \right\rvert = 3\lvert y\rvert \cdot \frac{x^2}{x^2 + y^2} \le 3\lvert y\rvert.

As $(x, y) \to (0, 0)$ , $3\lvert y\rvert \to 0$ . By the Squeeze Theorem, the limit is $0$ .

Polar coordinates package the same idea. With $x = r\cos\theta$ and $y = r\sin\theta$ , the condition $(x, y) \to (0, 0)$ becomes $r \to 0^+$ , for every $\theta$ . In the example above,

\frac{3x^2 y}{x^2 + y^2} = \frac{3r^3\cos^2\theta\sin\theta}{r^2} = 3r\cos^2\theta\sin\theta,

and since $\lvert 3\cos^2\theta\sin\theta\rvert \le 3$ , the expression is at most $3r$ in absolute value, which goes to $0$ . The bound must not depend on $\theta$ ; if the polar form still depends on $\theta$ after $r \to 0$ , the limit does not exist.

Tip

A quick heuristic at the origin: compare the total degree of the numerator with that of the denominator. In $\dfrac{3x^2y}{x^2 + y^2}$ the numerator has degree 3 and the denominator degree 2, so the limit is often $0$ . When the degrees match, as in $\dfrac{xy}{x^2 + y^2}$ , the limit usually fails. This is only a guide; always confirm with a squeeze or two paths.

Continuity

A function is continuous on a region if it is continuous at every point of the region. A piecewise definition such as

f(x, y) = \begin{cases} \dfrac{xy}{x^2 + y^2}, & (x, y) \ne (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases}

is continuous everywhere except the origin: along $y = x$ the values are $\tfrac{1}{2}$ , not $0$ , so the limit at the origin does not exist.