Limits of Functions of Two Variables

Two sides on a line

For a function of one variable, the limit as x tends to a asks what value f(x) gets close to when x is close to a, but not equal to it. On a number line x can come toward a from only two sides: from the left, through values below a, and from the right, through values above it.

So one-variable limits are checked from two sides. The limit exists when the left-hand limit and the right-hand limit are the same number, and that number is the limit.

Every path on a plane

A function of two variables takes a point (x, y) of the plane and gives a height f(x, y). Its limit as (x, y) tends to (a, b) asks what value f(x, y) gets close to when the point (x, y) is close to (a, b), but not equal to it. Close means that the distance between the two points is small.

On a plane a point can come toward (a, b) from any direction: along the x-axis, along the line y = x, along a parabola, or spiraling in. There are infinitely many paths. The limit is L only if f(x, y) gets close to L along every one of them, however (x, y) approaches.

Two paths that disagree

Take $f\left(x,y\right)=\frac{2xy}{x^2+y^2}$f(x, y) = 2xy/(x² + y²) near the origin. Substituting x = 0 and y = 0 gives $\frac{0}{0}$0/0, which says only that substitution fails. The behavior near (0, 0) has to be found along paths.

Along the x-axis, y = 0, so the top is 0 and f = 0 at every point except the origin. Along the line y = x, $f=\frac{2x^2}{x^2+x^2}=\frac{2x^2}{2x^2}=1$f = 2x²/(x² + x²) = 2x²/2x² = 1 at every point except the origin. At x = 0.1, 0.01 and 0.001 on those two lines, f is 0 and 1 each time.

Two paths into (0, 0) give two different values, 0 and 1. No single number L is close to both, so the limit does not exist. One disagreement is enough.

Along any line y = mx, $f=2m\frac{x^2}{x^2+m^2{x}^2}=\frac{2m}{1+m^2}$f = 2m x²/(x² + m² x²) = 2m/(1 + m²). The x cancels, so f has the same value at every point of the line, and that value depends on the gradient m. The line y = 2x gives $\frac{4}{5}$4/5, and the line y = −x gives −1.

The value of $f=\frac{2xy}{x^2+y^2}$f = 2xy/(x² + y²) along the line y = mx, plotted against the gradient m: $f=\frac{2m}{1+m^2}$f = 2m/(1 + m²). The marked points are the lines y = 0, y = x, y = −x and y = 2x, which give 0, 1, −1 and 0.8. Each line gives its own constant value, so the paths into the origin disagree.

A ray into the origin at angle $\theta$θ, over the same function. Along the ray $f=\sin 2\theta$f = sin 2θ at every distance, so at 30° it reads sin 60° = 0.87 all the way in. Along y = x, at 45°, it reads 1, and along y = −x, at 135°, it reads −1.

Agreement along lines is not enough

Two paths that agree prove nothing, and neither do all the straight lines together. Take $g\left(x,y\right)=\frac{x^{2}y}{x^4+y^2}$g(x, y) = x²y/(x⁴ + y²) near the origin.

Along the x-axis, g = 0. Along any line y = mx with m not 0, $g=m\frac{x^3}{x^4+m^2{x}^2}=m\frac{x}{x^2+m^2}$g = m x³/(x⁴ + m² x²) = m x/(x² + m²), which tends to 0 as x tends to 0. Along y = x, for example, g is about 0.099 at x = 0.1 and about 0.01 at x = 0.01. Every straight line gives 0.

Now approach along the parabola $y=x^2$y = x². Then $g=\frac{x^4}{x^4+x^4}=\frac{1}{2}$g = x⁴/(x⁴ + x⁴) = 1/2 at every point of it except the origin. A curved path gives ½, the straight lines give 0, so g has no limit at (0, 0).

The height of $g=\frac{x^{2}y}{x^4+y^2}$g = x²y/(x⁴ + y²) along two paths, plotted against x. The gold line is the path $y=x^2$y = x², where g = ½ at every point. The dashed curve is the path y = x, where $g=\frac{x}{x^2+1}$g = x/(x² + 1), which runs into 0. Neither path includes the origin, so both are drawn open there.

Showing that a limit exists

Paths can only disprove a limit. To prove one, find a bound that holds on every path at once. Take $h\left(x,y\right)=\frac{x^{2}y}{x^2+y^2}$h(x, y) = x²y/(x² + y²) near the origin.

Since $x^2\le x^2+y^2$x² ≤ x² + y², the fraction $\frac{x^2}{x^2+y^2}$x²/(x² + y²) is between 0 and 1. So h, which is that fraction times y, is never farther from 0 than y is: $-|y|\le h\left(x,y\right)\le |y|$−|y| ≤ h(x, y) ≤ |y|.

As (x, y) tends to (0, 0) along any path, y tends to 0, so both bounds tend to 0, and h is squeezed between them. The limit of h is 0. At (0.01, −0.02), for example, h = −0.004, which is inside the bound 0.02.

The usual mistakes

Substituting the point and giving $\frac{0}{0}$0/0, or 0, as the limit. The form $\frac{0}{0}$0/0 says only that substitution fails.

Testing one path, or the two axes, and calling the value found the limit. $\frac{2xy}{x^2+y^2}$2xy/(x² + y²) is 0 along both axes and has no limit at the origin.

Testing every straight line and calling it a proof. $\frac{x^{2}y}{x^4+y^2}$x²y/(x⁴ + y²) gives 0 along every line through the origin and ½ along the parabola $y=x^2$y = x².

A mixer tap

In the application below, the temperature of the water from a mixer tap is a function of the hot and cold flows, and it has no value when both are 0. Closing the tap along a line y = mx keeps the temperature fixed at $\frac{60+15m}{1+m}$(60 + 15m)/(1 + m), so two different mixes, closed down, give two different limits.