MATH U101 · Multivariable Calculus · Module 3

Functions of Several Variables: Domains, Level Curves, Limits & Continuity

Thomas §14.1–14.2, lectures L6–L7. The doorway to the rest of the course: every later module differentiates or integrates the functions defined here.

🎧 Narrated walkthroughs for this module

Finding limits (6.5 min, Tutorial 5 Q1, Q6–Q9) · When limits fail, and continuity (7.9 min, Tutorial 5 Q2–Q5, Q10). Both are sound-on and pause for your next move.

Start here

Nothing in this module is 12th-class material — §14.1's vocabulary (domain as a region, level curves, open/closed) is new to every student in the room, and §14.2 reuses the one-variable limit tools you drilled in Module 2 with one new twist (paths). There is no background gap to close here, only new words to learn precisely.

First time? Start with the Module 3 lesson (same material, taught slowly with a level-curve widget); this page is the compressed map for revision.

1 · Functions of two variables; domain and range §14.1

Intuition: a landscape — every point of the plane gets an altitude z = f(x, y). Formal: a rule assigning one real number to each point of a set D (the domain) in the plane; the set of values taken is the range. Unless stated, the domain is the natural domain: every point where the formula makes sense.

Square root: inside ≥ 0. Log: inside > 0. Fraction: denominator ≠ 0. Turn each into an inequality in x, y and sketch the region.
Range: ask what values the inside takes over the domain, then apply the outer function. (√(9−x²−y²) has range [0, 3], not [0, ∞).)

2 · Graphs, level curves, level surfaces

The graph z = f(x, y) is a surface. Its level curve at level c is the set f(x, y) = c in the plane — a contour line. "Describe the level curves" = name the curve family and how it changes with c, including the degenerate values (empty; a single point; the axes).

f(x, y)	Level curve f = c	Family
x² + y²	x² + y² = c	circles radius √c (c > 0); origin (c = 0); empty (c < 0)
y / x²	y = cx², x ≠ 0	parabolas through (but excluding) the origin; c = 0 → x-axis minus origin
xy	xy = c	hyperbolas, axes as asymptotes; c = 0 → the two axes
x² − y²	x² − y² = c	hyperbolas (left/right for c > 0, up/down for c < 0); c = 0 → lines y = ±x
√(x² + y²)	x² + y² = c²	circles radius c (c ≥ 0), evenly spaced — a cone

For w = f(x, y, z) there is no graph to draw; the set f = c is a level surface in space (spheres for x²+y²+z², planes for x+2y+3z).

3 · Interior, boundary, open, closed, bounded

Term	Test (with a small disk around the point)
Interior point	some disk around it lies entirely in the region
Boundary point	every disk around it meets both the region and its complement (the point itself may be in or out)
Open region	all its points are interior — contains none of its boundary
Closed region	contains all its boundary points
Bounded	lies inside some disk of finite radius; else unbounded

Recipe: boundary = the domain inequality turned into an equation. Strict (<, >, ≠) ⟹ boundary excluded ⟹ open. Non-strict (≤, ≥) ⟹ included ⟹ closed. Mixed ⟹ neither. The whole plane is both open and closed (no boundary).

Worked example 1The full §14.1 workout (Tutorial 4, Q5 pattern) for f(x, y) = √(9 − x² − y²) and g(x, y) = 1/(xy)

Domain of f: need 9 − x² − y² ≥ 0, i.e. x² + y² ≤ 9: the closed disk of radius 3.
Range of f: on the disk the inside runs from 9 (origin) down to 0 (rim), so √ of it runs over [0, 3].
Level curves: √(9 − x² − y²) = c ⟹ x² + y² = 9 − c²: circles of radius √(9 − c²) for 0 ≤ c < 3; the single point (0, 0) for c = 3; nothing for other c. (The graph is the upper hemisphere of radius 3.)
Boundary: the circle x² + y² = 9. It is included (≤), so the domain is closed; it fits in a disk, so bounded.
Domain of g: xy ≠ 0 — the plane with both axes removed.
Range of g: xy takes every nonzero real value, so does its reciprocal: (−∞, 0) ∪ (0, ∞).
Level curves: 1/(xy) = c ⟹ xy = 1/c: hyperbolas with the axes as asymptotes, one for every c ≠ 0; no level curve for c = 0.
Boundary: the two axes (every disk around an axis point contains points of the domain and points on the axis). None of it is in the domain, so open; points go off to infinity, so unbounded.

4 · Limits in two variables §14.2

Intuition: unchanged — outputs settle at L as the input point closes in on (a, b), from every direction and along every path. Formal: lim_{(x,y)→(a,b)} f(x, y) = L if for every ε > 0 there is a δ > 0 such that 0 < √((x−a)² + (y−b)²) < δ implies |f(x, y) − L| < ε. The limit laws (sum, product, quotient, power, root) all hold, and polynomials and rational functions have limits by substitution wherever defined.

5 · Showing a limit does not exist: the two-path test

If f approaches different values along two different paths into (a, b), the limit does not exist. Path ladder, cheapest first: axes (y = 0, x = 0) → lines y = mx → curves that balance the denominator's powers (x⁴ + y² ⟹ try y = kx²). A result that depends on m or k is a complete disproof.

The two-path test on f = xy/(x²+y²): every road into the origin must agree on the altitude — the axes say 0, the line y = x says ½, so no single limit exists.

Worked example 2Show lim_{(x,y)→(0,0)} xy/(x² + y²) does not exist

Substituting gives 0/0, so the limit needs investigating. Along the x-axis (y = 0): f(x, 0) = 0/x² = 0 → 0. Along the y-axis: also 0. The axes agree, so climb the ladder.
Along the line y = mx (x ≠ 0): f(x, mx) = mx²/(x² + m²x²) = m/(1 + m²) — constant along the line, and different for different m.
Concretely: along y = x (m = 1) the values are ½; along y = 0 they are 0. Two paths, two values ⟹ the limit does not exist.

Unpack this step

Cancelling x² in step 2 is legal because x ≠ 0 on the approach (the limit never looks at the point itself). The answer m/(1+m²) not depending on x means the function is constant along each line through the origin — the surface is a "ruled" cliff, and its height at the origin depends on which way you walk in.

6 · Showing a limit exists: algebra, polar, squeeze

Substitution when the formula is continuous at the point (§7).
Algebra for 0/0: factor and cancel, or multiply by the conjugate — the §2.4 tools unchanged. E.g. (x² − xy)/(√x − √y) → conjugate → x(√x + √y) → 0.
Polar at the origin: x = r cos θ, y = r sin θ, x² + y² = r²; show the expression → L as r → 0 by a bound that holds for all θ. If the r → 0 value depends on θ, the limit does not exist.
Squeeze: |f| ≤ (something → 0), typically using x²/(x² + y²) ≤ 1 or |x| ≤ √(x² + y²).

Worked example 3Prove lim_{(x,y)→(0,0)} 3x²y/(x² + y²) = 0, two ways

Suspect the value first. Along y = 0: 0. Along y = mx: 3mx³/((1 + m²)x²) = 3mx/(1 + m²) → 0. All paths so far say 0 — so try to prove 0 (a path check is never the proof).
Polar. 3x²y/(x² + y²) = 3r²cos²θ · r sin θ / r² = 3r cos²θ sin θ. Since |cos²θ sin θ| ≤ 1, |3r cos²θ sin θ| ≤ 3r → 0 as r → 0, regardless of θ. Hence the limit is 0.
Squeeze (no polar). |3x²y/(x² + y²)| = 3|y| · x²/(x² + y²) ≤ 3|y|, because a part over the whole is at most 1. As (x, y) → (0, 0), 3|y| → 0, so the function is squeezed to 0.
Sanity numbers (along y = x, x = 0.1, 0.01, 0.001): 0.15, 0.015, 0.0015 — shrinking like 3r/2, as the polar form predicts.

7 · Continuity

f is continuous at (a, b) iff f(a, b) is defined, the limit exists, and they are equal. Sums, products, quotients (nonzero denominator) and compositions of continuous functions are continuous; hence every formula built from polynomials, roots, exp, log, trig is continuous on its natural domain. Exam pattern: a formula patched by hand at one point (f(0, 0) = c) — check whether the limit exists (§5–6) and equals the patch. (x² − y²)/(x² + y²) patched at the origin: limit is 1 along y = 0, −1 along x = 0 ⟹ no patch works. x³/(x² + y²) patched with 0: polar gives r cos³θ → 0 ⟹ continuous.

Classic traps

1) A path proves existence. It never does — one path (or all straight lines) agreeing only tells you what the limit would have to be. 2) Stopping at the axes. If the axes agree, try y = mx; if lines agree, try the curve that balances the denominator (x⁴ + y² ⟹ y = kx²). 3) Range without the domain. √(9 − x² − y²) has range [0, 3]. 4) Open/closed by feel. Decide from the inequality: strict ⟹ open, non-strict ⟹ closed, mixed ⟹ neither. 5) Polar with a θ left over. If your r → 0 result still contains θ (e.g. cos 2θ), that's a disproof, not a limit.

Your minimal prerequisite kit for this module

Inequalities as regions: y ≥ x² is "above the parabola", x² + y² < 4 is "inside the circle of radius 2", xy > 0 is "first and third quadrants". Sketch, then shade.
Conics by sight: x² + y² = c circle; y = cx² parabola; xy = c, x² − y² = c hyperbolas; x²/a² + y²/b² = 1 ellipse. (Thomas §11.6 / any 11th-class conics chapter — a 30-minute skim.)
Where √ and ln live: √u needs u ≥ 0 and outputs ≥ 0; ln u needs u > 0 and outputs every real number.
The 0/0 toolbox from Module 2, Part A: factor/cancel, conjugate, sin u/u → 1, squeeze.
Polar conversion from Module 1: x = r cos θ, y = r sin θ, x² + y² = r², and |cos θ|, |sin θ| ≤ 1.

What to practise in Thomas

Exercise numbers are approximate for the 15th edition — check your copy. Prefer odd numbers (answers in the back).

Skill	Where	How many
Domain and range; sketch the domain	§14.1 exercises ~1–30 (domain/range/level-curve sets)	8–10 odd
Describe / sketch level curves for given levels	§14.1 exercises ~31–36 (and the matching-graph set)	4–6
Level surfaces of f(x, y, z)	§14.1 exercises ~37–48	3–4
Boundary; open/closed/neither; bounded	the (d)–(e) parts of the §14.1 domain exercises	every one you do
Limits by substitution and algebra	§14.2 exercises ~1–30	8–10 odd
Continuity: where is f continuous?	§14.2 exercises ~31–40	4–5
Two-path test (lines, then curves)	§14.2 exercises ~41–48	all — this is the quiz favourite
Polar-coordinate limits	§14.2 exercises ~57–64	4–6

Tutorial 4, Q5 is the §14.1 workout on three functions (y/x², xy, ln(x² + y² − 1)); Worked example 1 above is the same five-part pattern on different functions — do the tutorial cold, with the example closed.