Multivariable Limits and Continuity, multivariable and vector calculus lesson

On a line you can approach a point from only two sides. In the plane you can come in along any line, along a parabola, or spiraling inward, and a function may behave differently on each route. A limit in several variables has to account for all of them at once. The way to do that is to stop thinking about routes and measure only distance.

Key terms

The epsilon-delta definition using distance

Definition Let $f$ be a function with domain $D \subseteq \mathbb{R}^2$ , and let $(a, b)$ be a point such that every open disk about $(a, b)$ contains points of $D$ other than $(a, b)$ itself. Such a point is called a limit point of $D$ , and it need not belong to $D$ . We write

\lim_{(x, y) \to (a, b)} f(x, y) = L

if for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\lvert f(x, y) - L \rvert < \varepsilon$ for every $(x, y)$ in $D$ with

0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta.

For three variables replace the distance by $\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}$ . When the limit exists it is unique.

In plain words However thin an output band around $L$ you choose, there is a small disk around the input point such that every allowed input inside it, other than the center, lands in that band.

Example To prove that $x + y \to 3$ as $(x, y) \to (1, 2)$ , write $d$ for the distance from $(x, y)$ to $(1, 2)$ . Each of $\lvert x - 1 \rvert$ and $\lvert y - 2 \rvert$ is at most $d$ , so

\begin{aligned} \lvert x + y - 3 \rvert &\le \lvert x - 1 \rvert + \lvert y - 2 \rvert \\[0.4em] &\le 2d. \end{aligned}

Given $\varepsilon > 0$ , choose $\delta = \varepsilon / 2$ . Then $d < \delta$ gives $\lvert x + y - 3 \rvert < 2\delta = \varepsilon$ .

All approaches

Definition The definition controls every point of $D$ within distance $\delta$ , regardless of how it was reached. Consequently, if the limit of $f$ at $(a, b)$ is $L$ and $C$ is any subset of $D$ that has $(a, b)$ as a limit point, the limit of $f$ restricted to $C$ is also $L$ . In particular, if $\mathbf{r}(t)$ is a curve with $\mathbf{r}(t) \in D$ and $\mathbf{r}(t) \ne (a, b)$ for $t$ near $t_0$ , $t \ne t_0$ , and $\mathbf{r}(t) \to (a, b)$ as $t \to t_0$ , then

\lim_{t \to t_0} f(\mathbf{r}(t)) = L.

In plain words A true limit is the same along every road into the point, straight or curved, fast or slow.

Example The limit of $f(x, y) = x^2 + y^2$ at the origin is $0$ , since $f$ is the squared distance. Along the spiral $\mathbf{r}(t) = \langle t \cos(1/t), t \sin(1/t) \rangle$ with $t \to 0^+$ , which circles the origin infinitely often, the values are $f(\mathbf{r}(t)) = t^2$ , and these tend to $0$ as the definition promises.

Path tests can disprove a limit

Definition Suppose $C_1$ and $C_2$ are two paths in $D$ that approach $(a, b)$ , each having $(a, b)$ as a limit point, for example curves in $D$ that tend to $(a, b)$ without passing through it. If $f(x, y) \to L_1$ along $C_1$ and $f(x, y) \to L_2$ along $C_2$ with $L_1 \ne L_2$ , then the limit of $f$ at $(a, b)$ does not exist. The same conclusion holds if the limit along a single path fails to exist.

In plain words One road gives one answer and another road gives a different answer, so there is no single answer.

Example Let

\begin{gathered} f(x, y) = \frac{xy}{x^2 + y^2}, \\[0.4em] (x, y) \ne (0, 0). \end{gathered}

Along the line $y = mx$ with $x \ne 0$ ,

f(x, mx) = \frac{m x^2}{x^2 + m^2 x^2} = \frac{m}{1 + m^2}.

Along the $x$ -axis ( $m = 0$ ) the values are $0$ , and along $y = x$ ( $m = 1$ ) they are $\frac{1}{2}$ . So the limit at the origin does not exist.

Agreeing paths do not prove a limit

Definition Finding the same limit $L$ along some paths, even along every line through $(a, b)$ , does not show that the limit is $L$ . A path test can only disprove. To prove a limit you need an argument that covers every point of $D$ near $(a, b)$ at once, such as the definition, a squeeze estimate or continuity.

In plain words Checking a thousand roads still leaves infinitely many unchecked, and the bad one might be curved.

Example Consider

\begin{gathered} f(x, y) = \frac{x^2 y}{x^4 + y^2}, \\[0.4em] (x, y) \ne (0, 0). \end{gathered}

Along the $y$ -axis $f = 0$ . Along a line $y = mx$ , including the $x$ -axis $m = 0$ ,

\begin{aligned} f(x, mx) &= \frac{m x^3}{x^4 + m^2 x^2} \\[0.4em] &= \frac{m x}{x^2 + m^2} \to 0. \end{aligned}

So every line through the origin gives $0$ . But along the parabola $y = x^2$ ,

f(x, x^2) = \frac{x^4}{2x^4} = \frac{1}{2}.

The limit at the origin does not exist.

Squeeze estimates

Definition Let $(a, b)$ be as in the definition of a limit. If there is a function $g$ with

\lvert f(x, y) - L \rvert \le g(x, y)

for all $(x, y)$ in $D$ in some punctured disk about $(a, b)$ , and $g(x, y) \to 0$ as $(x, y) \to (a, b)$ , then $f(x, y) \to L$ . The most useful bounds are functions of the distance $r = \sqrt{(x - a)^2 + (y - b)^2}$ alone, built from the inequalities $\lvert x - a \rvert \le r$ and $\lvert y - b \rvert \le r$ .

In plain words Trap the error between $0$ and something you know shrinks to $0$ . A bound in terms of distance alone works for every approach at once.

Example Let

\begin{gathered} f(x, y) = \frac{x^2 y}{x^2 + y^2}, \\[0.4em] (x, y) \ne (0, 0). \end{gathered}

Since $x^2 \le x^2 + y^2$ , the fraction $\frac{x^2}{x^2 + y^2}$ is at most $1$ , so $\lvert f(x, y) \rvert \le \lvert y \rvert \le r$ . As $r \to 0$ the bound goes to $0$ , so the limit is $0$ . In epsilon-delta form, $\delta = \varepsilon$ works.

Continuity relative to a domain

Definition Let $f$ have domain $D$ and let $(a, b)$ be a point of $D$ . The function $f$ is continuous at $(a, b)$ if for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\lvert f(x, y) - f(a, b) \rvert < \varepsilon$ for every $(x, y)$ in $D$ with $\sqrt{(x - a)^2 + (y - b)^2} < \delta$ . Only points of $D$ are tested, so at a boundary point of $D$ that belongs to $D$ the inputs come from inside $D$ only. If $(a, b)$ is an isolated point of $D$ , meaning a point of $D$ that is not a limit point of $D$ , then a small enough disk contains no other point of $D$ and $f$ is automatically continuous there. When $(a, b)$ is a limit point of $D$ the definition says exactly that

\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b).

The function is continuous on $D$ if it is continuous at every point of $D$ . Sums, differences and products of continuous functions are continuous, quotients are continuous where the denominator is not zero, and if $f$ is continuous at $(a, b)$ and a function $g$ of one variable is continuous at $f(a, b)$ , then the composition $g(f(x, y))$ is continuous at $(a, b)$ . So polynomials in $x$ and $y$ are continuous on $\mathbb{R}^2$ , and rational functions are continuous on their domains.

In plain words A continuous function has no jumps or holes within its own domain, and its limit at each such point is simply its value there.

Example The function $\ln(x + y)/(xy)$ is built from continuous pieces and is continuous wherever $x + y > 0$ and $xy \ne 0$ , which includes $(1, 2)$ . So its limit there is found by substitution,

\lim_{(x, y) \to (1, 2)} \frac{\ln(x + y)}{xy} = \frac{\ln 3}{2}.

The function $\sqrt{1 - x^2 - y^2}$ is continuous on its domain, the closed unit disk, including the boundary circle, where only inputs from the disk count. In contrast, setting $f(0, 0) = 0$ for $f(x, y) = xy/(x^2 + y^2)$ gives a function on all of $\mathbb{R}^2$ that is not continuous at the origin, because its limit there does not exist.

Check yourself

Show that the limit of $\dfrac{x^2 - y^2}{x^2 + y^2}$ as $(x, y) \to (0, 0)$ does not exist.

Show the answer

Along the $x$ -axis the function equals $x^2/x^2 = 1$ , and along the $y$ -axis it equals $-y^2/y^2 = -1$ . Two paths give different limits, so the limit does not exist.

Find the limit of $\dfrac{xyz}{x^2 + y^2 + z^2}$ as $(x, y, z) \to (0, 0, 0)$ .

Show the answer

Let $r = \sqrt{x^2 + y^2 + z^2}$ . Each of $\lvert x \rvert$ , $\lvert y \rvert$ , $\lvert z \rvert$ is at most $r$ , so $\lvert xyz \rvert \le r^3$ and

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

The bound tends to $0$ , so the limit is $0$ by the squeeze estimate.

Let $f(x, y) = \dfrac{3x^2 y}{x^2 + y^2}$ for $(x, y) \ne (0, 0)$ . Which value of $f(0, 0)$ makes $f$ continuous on $\mathbb{R}^2$ ?

Show the answer

Away from the origin $f$ is a rational function with nonzero denominator, so it is continuous there. At the origin, $\lvert f(x, y) \rvert \le 3 \lvert y \rvert \le 3r$ , so the limit is $0$ . Setting $f(0, 0) = 0$ makes $f$ continuous everywhere.

Multivariable Limits and Continuity

Key terms

The epsilon-delta definition using distance

All approaches

Path tests can disprove a limit

Agreeing paths do not prove a limit

Squeeze estimates

Continuity relative to a domain

Check yourself

How this lesson connects