Difficulty: Intermediate | Prerequisites: Single-variable calculus (limits, continuity from Calc I/II), basic familiarity with functions of two variables.
This topic sits at the transition point between single-variable and multivariable calculus. You are moving from thinking about functions on a number line to functions on a plane, and the rules for continuity get more subtle. If you are comfortable with limits and the epsilon-delta definition from Calc I, you have the right foundation. If not, revisit those first. The reason this matters: continuity underpins nearly every major theorem you will meet in Calc III (partial derivatives, the chain rule for multivariable functions, optimisation with Lagrange multipliers). Getting it right here saves confusion later.
A function is continuous at a point when its limit there equals its value. For single-variable functions, you only need to check approach from the left and right. For multivariable functions, you must worry about every possible path of approach, which makes continuity far harder to verify and opens the door to subtle traps where a function looks continuous along some paths but is not continuous overall.
Continuous function (single variable)
A function g(x) is continuous at a point x₀ if three conditions hold: g(x₀) is defined, the limit of g(x) as x approaches x₀ exists, and that limit equals g(x₀). In simple terms, you can draw the graph through x₀ without lifting your pen.
Continuous function (two variables)
A function f(x, y) is continuous at a point (x₀, y₀) if f(x₀, y₀) is defined, the limit of f(x, y) as (x, y) approaches (x₀, y₀) exists, and that limit equals f(x₀, y₀). Think of it as: the surface has no holes or jumps at that point, no matter which direction you walk toward it from.
Limit (single variable)
The value that g(x) approaches as x gets arbitrarily close to x₀. For the limit to exist, the left-hand limit and right-hand limit must agree. In simple terms, the function settles on the same number whether you sneak up from the left or the right.
Limit (multivariable)
The value that f(x, y) approaches as (x, y) gets arbitrarily close to (x₀, y₀) along every possible path. Think of it as: no matter which direction you approach from across the plane, the function homes in on the same number.
Path of approach
A curve in the xy-plane along which you let (x, y) tend toward (x₀, y₀). Common choices include y = x, y = x², y = -2x, and so on. In simple terms, it is the route you take to walk toward the point on the surface.
For g(x) to be continuous at x₀, two things must line up:
The left-hand limit equals the right-hand limit: lim(x→x₀⁺) g(x) = lim(x→x₀⁻) g(x).
That common limit equals the function value: lim(x→x₀) g(x) = g(x₀).
Because x can only approach x₀ from two directions (left and right), checking continuity is relatively straightforward. You compute the one-sided limits, confirm they match, and confirm the function is defined there with the same value.
For f(x, y) to be continuous at (x₀, y₀), the same three conditions apply: the function must be defined at the point, the limit must exist, and the limit must equal the function value.
The critical difference is what "the limit exists" means. In single-variable calculus, there are only two directions. In two variables, (x, y) can approach (x₀, y₀) from infinitely many directions and along infinitely many curves. The limit exists only if f(x, y) approaches the same value along every one of those paths.
This is why you cannot prove continuity at a point just by checking a few paths. You can, however, disprove it by finding even one path that gives a different limit value.
The lab explored a function h(x, y) and tested its continuity at the origin by substituting specific paths:
y = x (the diagonal). h(x, x) approached 0 as x→0. Continuous along this path.
y = -2x. h(x, -2x) also approached 0. Continuous along this path.
y = x² (the parabola). h(x, x²) equalled 1 for all x ≠ 0. Since the function value at the origin is 0 and the limit along this path is 1, the function is not continuous along y = x².
y = -x². Same issue as y = x².
The function h(x, y) was defined at (0, 0) with value 0. Along straight lines through the origin the limit was 0, matching the function value. But along the parabolic path y = x², the limit was 1 – a different value.
In one variable, you have two directions to check. In two variables, you have infinitely many paths. A function can be continuous along every straight line through a point yet discontinuous along a curved path. That is precisely what the lab demonstrated: h(x, y) passed the straight-line tests but failed along y = x². This means checking a handful of paths can never prove continuity; it can only disprove it. To prove continuity, you typically need an epsilon-delta argument or the squeeze theorem that covers all paths at once.
Single-variable continuity condition:
lim(x→x₀⁺) g(x) = lim(x→x₀⁻) g(x) = g(x₀)
Multivariable continuity condition:
lim((x,y)→(x₀,y₀)) f(x, y) = f(x₀, y₀)
where the limit must hold along every path of approach.
Path substitution technique:
Replace y with a function of x (e.g. y = mx, y = x², y = x³) to reduce f(x, y) to a single-variable function p(x) = f(x, g(x)). Then check lim(x→x₀) p(x). If any two paths yield different limits, the multivariable limit does not exist and f is discontinuous at that point.
Continuity matters whenever a model needs smooth, predictable behaviour. Temperature distributions across a metal plate are modelled by continuous functions of two variables, and engineers rely on that continuity to guarantee that heat flows smoothly without sudden jumps. In computer graphics, a surface that is discontinuous will show visible seams or tears in the rendered image. Any time you see a smooth 3D surface in a game or simulation, multivariable continuity is doing the work behind the scenes.
"If the limit exists along every straight line, the function is continuous." It is not. The lab showed h(x, y) has the same limit along every line through the origin, yet the limit along the parabola y = x² is different. Straight-line checks are necessary but not sufficient.
"Continuity along individual paths proves overall continuity." No number of individual path checks can prove continuity. You can only disprove it (by finding two paths with different limits). Proving continuity requires a general argument that covers all paths at once.
"A function that is defined at a point is automatically continuous there." Being defined is only one of the three conditions. The limit must also exist and equal the function value.
"Multivariable limits work the same as single-variable limits, just with more algebra." The conceptual leap is significant. In one variable, there are two directions. In two variables, there are infinitely many paths, including curves of every shape.
⚠️ Exam questions commonly give a piecewise function of two variables and ask whether it is continuous at the origin. The standard approach: test several paths (y = x, y = mx, y = x²). If two give different limits, you are done; if they all agree, note that you have not proven continuity and state what further argument would be needed.
⚠️ Know the three conditions for continuity cold. A question that asks "is f continuous at (a, b)?" is really asking you to verify all three, not just compute a limit.
⚠️ The parabolic path y = x² is a favourite exam trap. Many functions that are continuous along every straight line through the origin fail along y = x². Always test at least one curved path.
True or false: If lim(x→0) g(x) exists but g(0) is undefined, g is continuous at x = 0.
True or false: To prove a two-variable function is continuous at a point, it is enough to show the limit is the same along y = x, y = -x, and y = 0.
Fill in the blank: A function f(x, y) is continuous at (x₀, y₀) if f(x₀, y₀) is defined, the limit ______, and the limit equals ______.
True or false: Finding two paths with different limits proves the multivariable limit does not exist.
Fill in the blank: In the lab, h(x, x²) equalled ______ for all x ≠ 0, while h(0, 0) = ______, which showed discontinuity along that path.
Q: State the three conditions for f(x, y) to be continuous at (a, b).
A: (1) f(a, b) is defined. (2) lim((x,y)→(a,b)) f(x, y) exists. (3) That limit equals f(a, b).
Q: A function f(x, y) has the same limit along y = x, y = -x, y = 2x, and y = -3x as (x, y)→(0, 0). Can you conclude f is continuous at the origin?
A: No. All of those are straight-line paths. A function can have the same limit along every straight line yet a different limit along a curved path such as y = x². You would need an argument covering all paths (e.g. epsilon-delta or squeeze theorem) to conclude continuity.
Q: Suppose h(x, y) = 2x²y / (x⁴ + y²) for (x, y) ≠ (0, 0) and h(0, 0) = 0. Test continuity at the origin by evaluating h along y = x and y = x².
A: Along y = x: h(x, x) = 2x³ / (x⁴ + x²) = 2x / (x² + 1) → 0 as x→0. Along y = x²: h(x, x²) = 2x²·x² / (x⁴ + x⁴) = 2x⁴ / 2x⁴ = 1 for all x ≠ 0. Since the limit along y = x² is 1 ≠ 0 = h(0, 0), the function is not continuous at the origin.
Q: Why is checking continuity harder in two variables than in one?
A: In one variable, there are only two directions of approach (left and right). In two variables, there are infinitely many paths, including straight lines, parabolas, spirals, and arbitrary curves. Continuity requires the limit to be the same along all of them, which no finite set of path checks can guarantee.
Q: In the lab, which path revealed that h(x, y) was not continuous at the origin, and what was the limit along that path?
A: The path y = x² (the parabolic path). Along that path, h(x, x²) = 1 for all x ≠ 0, while h(0, 0) = 0, so the limit did not match the function value.
This material connects directly to partial derivatives: a function must be continuous for many derivative-based results to hold, but having partial derivatives at a point does not guarantee continuity there (another common exam question). It also sets the stage for differentiability of multivariable functions, where the bar is even higher than continuity. Later, when you reach the multivariable chain rule or Lagrange multipliers, continuity of the functions involved is a standing assumption you will need to verify.
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