Limits and continuity

A point in the plane can be approached from infinitely many directions and along infinitely many curves, so a limit in several variables has far more ways to fail than a limit on the real line, where there are only a left and a right.

The previous lesson ended with the tool this one needs: the distance from $(x, y)$ to $(a, b)$ is $\sqrt{(x - a)^{2} + (y - b)^{2}}$ , whatever direction separates them. This lesson uses that number to define limits and continuity in two variables, shows how a limit can depend on the path, and gives a reliable way to prove that one exists. It assumes the one variable limit from Calculus I.

Closeness is distance

In one variable, $lim_{x \to a} f (x) = L$ is a tolerance game: name any tolerance $ε$ on the output, and some tolerance $δ$ on the input keeps $f (x)$ within it. The several variable version keeps the game and changes only how the input tolerance is measured. Instead of $| x - a |$ , use the distance to the point:

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

means that for every $ε > 0$ there is a $δ > 0$ such that $0 < \sqrt{(x - a)^{2} + (y - b)^{2}} < δ$ forces $| f (x, y) - L | < ε$ .

Geometrically, the points within $δ$ of $(a, b)$ form an open disc, and the condition $0 <$ removes its centre. So the limit is a promise about a punctured disc: every point in it, in every direction, has a value within $ε$ of $L$ . The value at $(a, b)$ itself plays no part and need not exist, as at the hole of $x y / (x^{2} + y^{2})$ .

Producing a $δ$ works the same way as before, with one new habit: bound everything by the distance $d$ . Two facts do most of the work. Each coordinate difference is at most the distance, $| x - a | \leq d$ and $| y - b | \leq d$ , since a leg of a right triangle is never longer than its hypotenuse. And a dot product is at most the product of lengths, $| 𝐮 \cdot 𝐯 | \leq | 𝐮 | | 𝐯 |$ , since $| \cos θ | \leq 1$ .

Take $lim_{(x, y) \to (0, 0)} (3 x + 4 y) = 0$ . The quantity $3 x + 4 y$ is the dot product of $⟨ 3, 4 ⟩$ with $⟨ x, y ⟩$ , so $| 3 x + 4 y | \leq 5 d$ , and $δ = ε / 5$ works. The bound is sharp: at distance $0.01$ in the direction $⟨ 3, 4 ⟩$ , the point $(0.006, 0.008)$ gives $0.018 + 0.032 = 0.05$ , exactly $5 \times 0.01$ .

Every path must agree

If the limit is $L$ , then every point of the punctured disc has its value within $ε$ of $L$ , and in particular every point of any path that runs into $(a, b)$ . So the values along every path must tend to $L$ : along the $x$ axis, along any line, along a parabola, along a spiral. This gives the standard way to show that a limit does not exist. Find two paths into the point along which the function tends to different values, or one path along which it tends to nothing, and no $L$ can satisfy the definition.

This is the left and right limit test of one variable, with infinitely many sides. It runs in one direction only: agreement along two paths, or ten, or every line, proves nothing, because a path not yet tried may disagree. Paths disprove; they never prove.

A limit that depends on the direction

Take $f (x, y) = \frac{x y}{x^{2} + y^{2}}$ , defined everywhere except the origin. Along the $x$ axis, $y = 0$ and the function is $0$ at every point, so it tends to $0$ . Along the $y$ axis it is also $0$ . Along the diagonal $y = x$ it is $x^{2} / 2 x^{2} = \frac{1}{2}$ at every point, however close to the origin. Two paths, two different values, and the limit does not exist.

All the lines at once tell the whole story. Along $y = m x$ with $x \neq 0$ ,

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

which does not depend on $x$ at all: the function is constant along each line through the origin, with the constant set by the slope. The slope $m = 2$ gives $0.4$ , $m = - 1$ gives $- \frac{1}{2}$ , and as $m$ runs over the real numbers the constant covers every value from $- \frac{1}{2}$ to $\frac{1}{2}$ . So every punctured disc around the origin, however small, contains points where $f = \frac{1}{2}$ and points where $f = - \frac{1}{2}$ , and no tolerance smaller than $\frac{1}{2}$ can be met around any candidate $L$ .

Polar coordinates make the same point in one line. Writing $x = r \cos θ$ and $y = r \sin θ$ , where $r$ is the distance to the origin and $θ$ the angle of the direction, gives $f = \cos θ \sin θ = \frac{1}{2} \sin 2 θ$ . The distance $r$ has cancelled completely, so the value depends only on the direction. The level curves of $f$ are the lines through the origin, all crowding into the hole with every height from $- \frac{1}{2}$ to $\frac{1}{2}$ .

Example. Show that $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ has no limit at the origin.

Along $y = m x$ the value is $\frac{x^{2} - m^{2} x^{2}}{x^{2} + m^{2} x^{2}} = \frac{1 - m^{2}}{1 + m^{2}}$ , again constant on each line. The $x$ axis, $m = 0$ , gives $1$ . The line $y = x$ gives $0$ . The $y$ axis, which is not of the form $y = m x$ and must be checked on its own, has $x = 0$ and value $- y^{2} / y^{2} = - 1$ . Three different values along three paths, so there is no limit. In polar coordinates the function is $\cos^{2} θ - \sin^{2} θ = \cos 2 θ$ , which again has lost $r$ .

Now you. Find the value that $\frac{2 x y}{x^{2} + 4 y^{2}}$ approaches along the line $y = m x$ , and use it to show that the function has no limit at the origin.

Answer

Along $y = m x$ the value is $\frac{2 m x^{2}}{x^{2} + 4 m^{2} x^{2}} = \frac{2 m}{1 + 4 m^{2}}$ . The $x$ axis gives $0$ and the line $y = \frac{1}{2} x$ gives $\frac{1}{2}$ , so two paths disagree and the limit does not exist.

Every line is not enough

After the examples above it is tempting to check all the lines through the point and, if they agree, declare the limit. The function

g (x, y) = \frac{x^{2} y}{x^{4} + y^{2}}

shows why that fails. On the $x$ axis and the $y$ axis it is $0$ . Along $y = m x$ with $m \neq 0$ ,

g (x, m x) = \frac{m x^{3}}{x^{4} + m^{2} x^{2}} = \frac{m x}{x^{2} + m^{2}}

and as $x \to 0$ the numerator goes to $0$ while the denominator goes to $m^{2}$ , so the limit is $0$ . Every line into the origin gives $0$ . Numerically, along $y = x$ the values at $x = 0.1$ , $0.01$ and $0.001$ are $0.0990$ , $0.0100$ and $0.0010$ , heading steadily for zero.

Now approach along the parabola $y = x^{2}$ instead:

g (x, x^{2}) = \frac{x^{4}}{x^{4} + x^{4}} = \frac{1}{2}

at every point, however close. At $(0.01, 0.0001)$ the value is exactly $0.5$ . So there are points with $g = \frac{1}{2}$ in every punctured disc around the origin, and the limit does not exist, even though it is $0$ along every straight line. Along $y = k x^{2}$ the value is $k / (1 + k^{2})$ : $g$ is constant on parabolas the way $x y / (x^{2} + y^{2})$ is constant on lines.

Straight lines are simply the wrong family of paths for this function, and no finite list of families is always the right one. To prove a limit exists, something must control every point of a punctured disc at once.

Polar coordinates and the squeeze

The tool that controls a whole disc is a bound that depends on the distance alone. Suppose that near $(a, b)$

| f (x, y) - L | \leq B (d)

where $d$ is the distance to $(a, b)$ and $B (d) \to 0$ as $d \to 0$ . Given $ε$ , choose $δ$ so that $B (d) < ε$ whenever $d < δ$ , which the one variable limit of $B$ guarantees. Then every point of the punctured disc of radius $δ$ has $| f - L | < ε$ , whatever its direction. This is the squeeze theorem of Calculus I, with $f$ trapped between $L - B (d)$ and $L + B (d)$ , and both walls closing in on $L$ as the distance shrinks.

At the origin, polar coordinates are the natural way to find such a bound, because $r$ is the distance. Substitute $x = r \cos θ$ , $y = r \sin θ$ , simplify, and try to write $| f - L |$ as at most something in $r$ alone. Since $| \cos θ | \leq 1$ and $| \sin θ | \leq 1$ , any leftover trigonometric factors can be bounded by $1$ . The one requirement is that the final bound contain no $θ$ : a bound that shrinks for each fixed direction but not uniformly across all of them proves nothing.

Example. Prove that $\frac{3 x^{2} y}{x^{2} + y^{2}} \to 0$ as $(x, y) \to (0, 0)$ , and find a $δ$ for each $ε$ .

In polar coordinates the numerator is $3 r^{3} \cos^{2} θ \sin θ$ and the denominator is $r^{2}$ , so the function equals $3 r \cos^{2} θ \sin θ$ . Its size is at most $3 r$ , since the trigonometric factors are at most $1$ . The bound $B (r) = 3 r$ contains no $θ$ and tends to $0$ , so the limit is $0$ , and $δ = ε / 3$ works. As a check, at $(0.001, 0.001)$ the function is $0.003 \times 10^{- 6} / (2 \times 10^{- 6}) = 0.0015$ , and the bound there is $3 \times 0.001414 = 0.0042$ .

Now you. Prove that $\frac{x^{2} y^{2}}{x^{2} + y^{2}} \to 0$ at the origin, and give a $δ$ that works for $ε = 0.0001$ .

Answer

In polar coordinates the function is $r^{2} \cos^{2} θ \sin^{2} θ$ , at most $r^{2}$ with no $θ$ left, so the limit is $0$ . Any $δ$ with $δ^{2} \leq 0.0001$ works, for instance $δ = 0.01$ .

Counting degrees predicts the outcome. In $x y / (x^{2} + y^{2})$ the numerator and denominator both have degree $2$ , the powers of $r$ cancel exactly, and a pure function of $θ$ is left, which is what makes the value depend on the direction. In the example the numerator has degree $3$ , one spare power of $r$ survives, and it drives the whole expression to zero. The rule is a guide, not a theorem, and $g = x^{2} y / (x^{4} + y^{2})$ misleads it. In polar coordinates

g = \frac{r \cos^{2} θ \sin θ}{r^{2} \cos^{4} θ + \sin^{2} θ}

and for each fixed $θ$ with $\sin θ \neq 0$ this tends to $0$ as $r \to 0$ . But when $\sin θ$ is as small as $r \cos^{2} θ$ , the denominator shrinks to the size of $r^{2}$ and the fraction reaches $\frac{1}{2}$ ; those are the points of the parabola $y = x^{2}$ . No bound in $r$ alone exists, which is exactly the failure the requirement above rules out.

Where continuous functions come from

A function $f$ of two variables is continuous at $(a, b)$ when $f (a, b)$ is defined and $lim_{(x, y) \to (a, b)} f (x, y) = f (a, b)$ : the limit exists and the function takes the value it predicts. It is continuous on a region when it is continuous at every point of it. Continuity is recognised on sight the same way as in one variable: by building functions from simple ones with operations that preserve it.

The building blocks are the constants and the two coordinate functions $x$ and $y$ . The function $x$ is continuous everywhere because $| x - a | \leq d$ , so $δ = ε$ works; likewise $y$ .

Sums, products and quotients follow from the limit laws, and their one variable proofs go through unchanged, because they only ever manipulate output tolerances and never mention how the input distance is measured. The product shows the pattern. If $f \to L$ and $g \to M$ , write

f g - L M = f (g - M) + M (f - L)

Near the point $f$ stays within $1$ of $L$ , so $| f | \leq | L | + 1$ , and both terms can be made as small as wanted by making $| g - M |$ and $| f - L |$ small, each of which some $δ$ achieves; take the smaller of the two $δ$ values. So products of continuous functions are continuous, and repeating the product and sum rules shows that every polynomial in $x$ and $y$ is continuous everywhere. A quotient is continuous wherever its denominator is not zero, so every rational function is continuous except where the denominator vanishes. That includes $x y / (x^{2} + y^{2})$ everywhere but the origin, where no value fills the hole.

Composition is the rule that reaches everything else. Suppose $u = g (x, y)$ is continuous at $(a, b)$ , and $h$ is a function of one variable continuous at $c = g (a, b)$ . Given $ε$ , continuity of $h$ provides an $η$ such that $| u - c | < η$ keeps $| h (u) - h (c) | < ε$ . Continuity of $g$ then provides a $δ$ such that $d < δ$ keeps $| g (x, y) - c | < η$ . Chaining the two, $d < δ$ keeps $| h (g (x, y)) - h (g (a, b)) | < ε$ . So $e^{x y}$ , $\sin (x^{2} + y)$ , $\ln (x - y)$ on the half-plane $x > y$ , and $\sqrt{9 - x^{2} - y^{2}}$ on its disc are all continuous, since each is a continuous function of one variable applied to a polynomial.

The consequence is the same as in Calculus I: most limits are evaluated by substitution. Where nothing divides by zero the limit is the value, so $lim_{(x, y) \to (1, 2)} (x^{2} y + 3) / (x + y) = 5 / 3$ with no $ε$ in sight. The effort goes into the points where the formula breaks, and composition often handles those too.

Example. What value at the origin makes $F (x, y) = \frac{\sin (x^{2} + y^{2})}{x^{2} + y^{2}}$ continuous there?

Write $F = h (u)$ with $u = x^{2} + y^{2}$ and $h (u) = (\sin u) / u$ . Calculus I gives $\sin u / u \to 1$ as $u \to 0$ , so setting $h (0) = 1$ makes $h$ continuous at $0$ . The polynomial $u$ is continuous with $u (0, 0) = 0$ , so by the composition rule $F$ tends to $h (0) = 1$ , and defining $F (0, 0) = 1$ makes it continuous. At $(0.1, 0.2)$ , where $u = 0.05$ , the value is $0.99958$ .

Now you. What value at the origin makes $\frac{1 - \cos (x^{2} + y^{2})}{(x^{2} + y^{2})^{2}}$ continuous there?

Answer

With $u = x^{2} + y^{2}$ the function is $(1 - \cos u) / u^{2}$ . The Taylor polynomial $\cos u \approx 1 - \frac{1}{2} u^{2}$ gives the limit $\frac{1}{2}$ as $u \to 0$ , so the value is $\frac{1}{2}$ . At $(0.1, 0.2)$ the function is $0.49990$ .

One variable at a time, and what comes next

A tempting shortcut is to test continuity one variable at a time: freeze $y$ , check the resulting function of $x$ , then freeze $x$ and check the function of $y$ . The function $x y / (x^{2} + y^{2})$ , given the value $0$ at the origin, shows that this is not enough. Freeze $y = b$ with $b \neq 0$ and the result $b x / (x^{2} + b^{2})$ is a rational function of $x$ with a denominator that never vanishes, continuous everywhere. Freeze $y = 0$ and the result is the constant $0$ , also continuous. The same holds with the roles swapped. So the function is continuous in each variable separately at every point, including the origin, and yet it is not continuous at the origin, since along $y = x$ it stays at $\frac{1}{2}$ while its value there is $0$ .

The explanation is the theme of this lesson. Freezing a variable looks only along the two axes, two paths out of infinitely many, and paths can disprove continuity but never prove it. In three variables nothing changes but the distance formula: a ball replaces the disc.

Where continuity does hold, the next step is to measure slopes, and the idea that failed here as a test of continuity works well as a way of taking them. Freeze every variable but one, and the surface becomes a curve with an ordinary slope that Calculus I can compute. That is the partial derivative, the subject of the next lesson. It inherits the weakness exposed in this section: two slopes along the axes see only two directions, and $x y / (x^{2} + y^{2})$ will turn up again as a function that has them both at a point where it is not even continuous.