Continuity in Two Variables

No tears, holes or cliffs in the sheet.

Continuity on a curve

A function of one variable is continuous at x = a when its graph can be drawn through a without lifting the pencil. In symbols, three things hold: f(a) has a value, the limit of f(x) as x tends to a exists, and the two are equal.

The parabola y = x² is continuous everywhere. A function with a jump, a hole or a vertical asymptote fails at that point, because one of the three conditions breaks there.

The same three conditions on a surface

A function of two variables f(x, y) is continuous at the point (a, b) when three things hold: f(a, b) has a value, the limit of f(x, y) as (x, y) tends to (a, b) exists, and the limit equals f(a, b).

The middle condition is the one that changes in two variables: the limit has to be the same along every path into (a, b), so continuity says that every route to the point arrives at the height the surface actually has there.

Take f(x, y) = x² + 3xy at (1, 2). Its value is 1 + 6 = 7. At the nearby points (1.01, 2.01), (1.01, 1.99) and (0.99, 2.01) it is 7.11, 7.05 and 6.95. The heights close to (1, 2) are close to 7, on whichever side of the point they are. Since f is a polynomial, its limit at (1, 2) is found by substituting the point, so the limit is 7, the same as the value, and f is continuous there.

Which functions are continuous

Every polynomial in x and y, such as x² + 3xy or 5x³y − y², is continuous at every point of the plane. For a polynomial the limit at any point is found by substituting the point.

Sums, differences and products of continuous functions are continuous, and so is a continuous function of a continuous function: e^(xy) and sin(x + y) are continuous everywhere. A quotient is continuous wherever its denominator is not 0. So (x + y)/(x − y) is continuous at every point off the line y = x, and has no value on that line.

Where a surface fails

A function can fail to be continuous at a point in three ways. The first is a step. Let s(x, y) = 2 when y ≥ 0 and s(x, y) = 1 when y < 0. The surface is a flat shelf at height 2 over the upper half of the plane and a flat floor at height 1 over the lower half, with a cliff of height 1 along the x-axis. At any point (a, 0), the path coming down from above gives 2 and the path coming up from below gives 1, so there is no limit, and s is discontinuous at every point of the x-axis.

The second is a value that disagrees with the limit. The function x²y/(x² + y²) has no value at the origin, and its limit there is 0, because its size is at most the size of y. Give it the value 1 at the origin and the limit, 0, and the value, 1, disagree, so the function is discontinuous there. Give it the value 0 instead and it becomes continuous: this kind of discontinuity can be removed.

The third is a point with no limit at all. The function 2xy/(x² + y²) is 0 along the axes and 1 along the line y = x. It has no limit at the origin, so no value given to it there can make it continuous.

yz

The height of the step s along the y-axis, plotted against y. For negative y the height is 1; at y = 0 it jumps to 2 and stays there. The filled dot is the value s(0, 0) = 2, and the open dot is the height 1 that the path from below runs into.

xz

The function x²y/(x² + y²) along the line y = x, where it equals x/2, given the value 1 at the origin. The heights along the line run into 0, the open dot, but the value at the point is 1, the filled dot. The limit and the value disagree, so the function is discontinuous at the origin.

The usual mistakes

Checking only that f(a, b) has a value. The limit must also exist, and must equal that value.

Checking only that the limit exists. A limit of 5 with a value of 9 is a step of 4 at that one point, so the function is discontinuous there.

Reading 0/0 at a point as proof that the function is discontinuous for good. The formula has no value there, but the limit may still exist, and giving the function that value makes it continuous.

Two applications

In the first application below, the log mean temperature difference of a heat exchanger reads 0/0 wherever its two inputs are equal. Its limit at (10, 10) is 10, so defining its value there as 10 makes it continuous. In the second, the resistance of two rheostats joined in parallel is squeezed between 0 and a bound that tends to 0, so it is continuous at (0, 0) when given the value 0 there.

Worked example: A Heat Exchanger's Log Mean Temperature Difference: Its Value for Two End Differences, and Its Value When They Are Equal

Question In a heat exchanger, the hot stream is x degrees Celsius hotter than the cold stream at one end and y degrees hotter at the other. Engineers size the exchanger with the log mean temperature difference D(x, y) = x − yln x − ln y, which is defined for positive x and y except where x = y, when it reads 00. (a) Find D when the differences are 40°C and 10°C, and compare it with their ordinary average. (b) In some exchangers the two differences are equal. Find the limit of D(x, y) as (x, y) approaches (10, 10), and so the value D(10, 10) must take for D to be continuous there.

  1. 1.(a) D(40, 10) = 40 − 10ln 40 − ln 10 = 30ln 4 = 301.3863 ≈ 21.6°C. The ordinary average is 40 + 102 = 25°C, so the log mean is smaller, by about 3.4°C.

    102010203040x, with y held at 10D21.6average 25(a) D = 30 / ln 4 = 21.6, average 25
    102010203040x, with y held at 10D21.6average 25(a) D = 30 / ln 4 = 21.6, average 25
    (a) The log mean of 40°C and 10°C is 30ln 4 ≈ 21.6°C, below their ordinary average of 25°C.

  2. 2.(b) Write y = x(1 + u), where u = yx − 1. Then x − y = −xu and ln x − ln y = −ln(1 + u), so D = x × uln(1 + u) whenever u ≠ 0.

    102010203040x, with y held at 10D21.6average 25(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)
    102010203040x, with y held at 10D21.6average 25(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)
    (b) The curve is the cross-section y = 10. At x = 10 the formula reads 00, so the curve has a gap there.

  3. 3.As (x, y) approaches (10, 10) along any path, x approaches 10 and u = yx − 1 approaches 1010 − 1 = 0. The standard limit ln(1 + u)u → 1 as u → 0 gives uln(1 + u) → 1.

    102010203040x, with y held at 10D21.6average 25(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)x goes to 10, u goes to 0
    102010203040x, with y held at 10D21.6average 25(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)x goes to 10, u goes to 0
    Writing y = x(1 + u) turns D into x times a function of u alone, and u approaches 0 however (x, y) approaches (10, 10).

  4. 4.So D approaches 10 × 1 = 10 whichever way (x, y) approaches (10, 10): the limit is 10°C. Defining D(10, 10) = 10, the common difference, makes D continuous there; in the same way D(a, a) = a for every positive a.

    102010203040x, with y held at 10D21.6average 25D = 10(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)x goes to 10, u goes to 0u / ln(1 + u) goes to 1, so D goes to 10
    102010203040x, with y held at 10D21.6average 25D = 10(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)x goes to 10, u goes to 0u / ln(1 + u) goes to 1, so D goes to 10
    The limit is 10 along every path, so the value D(10, 10) = 10 closes the gap and makes D continuous there.

  5. 5.Check: D(10, 10.2) = −0.2ln 10 − ln 10.2 = −0.2−0.0198 ≈ 10.1, just above 10, as it should be for differences at or a little above 10°C.

    102010203040x, with y held at 10D21.6average 25D = 10(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)x goes to 10, u goes to 0u / ln(1 + u) goes to 1, so D goes to 10check: D(10, 10.2) = 10.1
    102010203040x, with y held at 10D21.6average 25D = 10(a) D = 30 / ln 4 = 21.6, average 25(b) y = x(1 + u): D = x × u / ln(1 + u)x goes to 10, u goes to 0u / ln(1 + u) goes to 1, so D goes to 10check: D(10, 10.2) = 10.1
    Close to (10, 10) the formula gives values close to 10: D(10, 10.2) ≈ 10.1.

Answer: (a) 30ln 4 ≈ 21.6°C, about 3.4°C below the ordinary average of 25°C; (b) the limit is 10°C, so D(10, 10) = 10 makes D continuous there

Common mistakes

  • Using the ordinary average, 25°C, in place of the log mean. Along the exchanger the difference changes exponentially, not in a straight line, so its average over the length is the log mean; the ordinary average overstates it and would make the exchanger too small.
  • Deciding that D has no limit at (10, 10) because it reads 00 there. 00 says only that substitution fails; the limit exists and is 10.

More functions of several variables problems, worked step by step →

Worked example: Two Rheostats Joined in Parallel: The Combined Resistance, and Its Limit as Both Are Turned Down to Zero

Question Two rheostats, which are variable resistors, are set to x and y ohms and joined in parallel. Their combined resistance is R(x, y) = xyx + y ohms for x ≥ 0 and y ≥ 0, except at (0, 0), where the formula reads 00. (a) Find the combined resistance when the rheostats are set to 6 ohms and 3 ohms. (b) Show that R(x, y) ≤ x + y4. Use it to find the limit of R as (x, y) approaches (0, 0), and so the value R(0, 0) must take for R to be continuous there.

  1. 1.(a) R(6, 3) = 6 × 36 + 3 = 189 = 2 ohms, less than either rheostat on its own, as a pair in parallel always is.

    346346x, ohmsy, ohmsR = 2(a) R = 18 / 9 = 2 ohms
    346346x, ohmsy, ohmsR = 2(a) R = 18 / 9 = 2 ohms
    (a) The settings 6 and 3 ohms give 2 ohms. The curve joins every pair of settings that gives 2 ohms.

  2. 2.(b) For the bound, start from (x − y)2 ≥ 0, which expands to x2 + y2 ≥ 2xy. Adding 2xy to both sides gives (x + y)2 ≥ 4xy.

    346346x, ohmsy, ohmsR = 2(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xy
    346346x, ohmsy, ohmsR = 2(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xy
    (b) Since a square is never negative, (x + y)2 ≥ 4xy for every pair of settings.

  3. 3.Dividing both sides by 4(x + y), which is positive everywhere except at (0, 0), gives xyx + y ≤ x + y4. So 0 ≤ R(x, y) ≤ x + y4 for every setting.

    346346x, ohmsy, ohmsR = 2(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xyso 0 ≤ R ≤ (x + y)/4
    346346x, ohmsy, ohmsR = 2(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xyso 0 ≤ R ≤ (x + y)/4
    So R ≤ x + y4: the curve R = 2 lies on or above the line x + y = 8 and touches it at (4, 4).

  4. 4.As (x, y) approaches (0, 0) along any path, x + y4 approaches 0, and R is squeezed between 0 and it. So the limit of R is 0 ohms, and defining R(0, 0) = 0, which is what the circuit gives when both rheostats are turned right down, makes R continuous there.

    346346x, ohmsy, ohmsR = 2R = 0(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xyso 0 ≤ R ≤ (x + y)/4(x + y)/4 goes to 0, so R goes to 0
    346346x, ohmsy, ohmsR = 2R = 0(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xyso 0 ≤ R ≤ (x + y)/4(x + y)/4 goes to 0, so R goes to 0
    The curves R = 1 and R = 0.5 shrink into the corner: close to (0, 0), R is squeezed to 0 from every direction.

  5. 5.Check: with both rheostats below 0.1 ohms, the bound gives R < 0.1 + 0.14 = 0.05 ohms. The settings 0.08 and 0.02 ohms give R = 0.00160.1 = 0.016 ohms, below that bound.

    346346x, ohmsy, ohmsR = 2R = 0(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xyso 0 ≤ R ≤ (x + y)/4(x + y)/4 goes to 0, so R goes to 0check: 0.08 and 0.02 give 0.016 ohms
    346346x, ohmsy, ohmsR = 2R = 0(a) R = 18 / 9 = 2 ohms(b) (x − y)2≥ 0, so (x + y)2≥ 4xyso 0 ≤ R ≤ (x + y)/4(x + y)/4 goes to 0, so R goes to 0check: 0.08 and 0.02 give 0.016 ohms
    With both settings below 0.1 ohms, R is below 0.05 ohms; setting R(0, 0) = 0 makes R continuous.

Answer: (a) 2 ohms; (b) R ≤ x + y4, so R approaches 0 ohms along every path, and R(0, 0) = 0 makes R continuous at (0, 0)

Common mistakes

  • Testing a few straight lines y = mx, finding 0 along each, and concluding that the limit is 0. Agreement along straight lines does not prove a limit, because a curved path could still give another value; the bound x + y4 covers every path at once.
  • Dividing by x + y at (0, 0) itself. The inequality holds only where x + y > 0; the point (0, 0) is exactly where the formula has no value, which is why its value there has to be chosen from the limit.

More functions of several variables problems, worked step by step →

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