The limit lim_{(x,y)→(a,b)} f(x, y) = L means f(x, y) approaches L as (x, y) approaches (a, b) along every possible path. This is fundamentally harder than single-variable limits: one-variable limits require checking only two directions (left and right), but in ℝ² there are infinitely many paths of approach. A function is continuous at (a, b) if the limit equals f(a, b). Showing a limit does not exist is typically done by finding two paths that give different limiting values.
Emphasize the path-dependence issue with a concrete example, such as f(x,y) = xy/(x²+y²) near the origin. Show that different approach paths (y=0, y=x, y=x²) give different limits. Then show how the squeeze theorem can establish that a limit does exist. The contrast between existence proofs and non-existence proofs builds the key skill.
In single-variable calculus, you defined lim_{x→a} f(x) = L by demanding that f(x) get arbitrarily close to L as x approaches a. The catch was simple: x can only approach a from the left or the right, so checking two directions was enough to confirm or refute a limit. In two variables, a point (a, b) in the plane is surrounded by infinitely many paths of approach — straight lines at every angle, parabolas, spirals, and more. A multivariable limit requires the function to converge to the same value L along every single one of these paths simultaneously.
This path-dependence is what makes multivariable limits genuinely harder. Consider f(x, y) = xy/(x² + y²) near the origin. Along the x-axis (y = 0), f = 0 for all x ≠ 0, so the limit from this path is 0. Along y = x, f = x²/(2x²) = 1/2 for all x ≠ 0, so the limit from this path is 1/2. Two paths give two different values, so the limit does not exist. This two-path test is the go-to strategy for showing a limit fails to exist: find two paths to (a, b) along which the function values approach different numbers.
But the two-path test cannot prove a limit exists. Even if the limit is 0 along every line y = mx through the origin, it might still fail along y = x². The path y = x² through the origin gives f(x, x²) = x · x²/(x² + x⁴) = x³/(x²(1 + x²)) = x/(1 + x²) → 0, so that path also gives 0 for this example — but this illustrates the need for caution, not a shortcut. To prove a limit exists, you need a genuine argument, typically the squeeze theorem: bound |f(x, y) − L| between 0 and a function of r = √(x² + y²) that goes to 0, exploiting the fact that any approach to (a, b) drives r → 0.
Continuity at (a, b) means the limit exists, equals f(a, b), and f is defined there — all three conditions simultaneously. For most elementary formulas (polynomials, rational functions away from their zeros, compositions of continuous functions), continuity holds everywhere the function is defined. The important exception is piecewise-defined functions, especially those with a special value assigned at a single point like the origin, where you must check whether the limit of the formula matches the assigned value. These continuity checks become essential prerequisites for partial derivatives, since differentiability requires the function to be well-behaved near a point in all directions.