Difficulty: Intermediate | Prerequisites: single-variable limits, derivatives, chain rule from Calculus I.
This block of material answers the question: how do limits, continuity, and derivatives work when your function has two or more inputs instead of one? The core surprise is that approaching a point along every straight line and getting the same answer still does not prove the limit exists, because there are infinitely many curved paths to check. That idea, together with tangent planes and the multivariable chain rule, forms the backbone of the entire course. If you are behind, make sure you are solid on single-variable limits and the ordinary chain rule before diving in.
To show a two-variable limit does not exist, find two paths to the point that give different limiting values. To show it does exist, you need an argument that covers all paths (e.g. squeeze theorem or polar coordinates). The tangent plane is the multivariable version of the tangent line, and linear approximation uses it to estimate function values near a known point. The chain rule extends naturally: multiply partial derivatives by the derivatives of the inner functions and add.
Multivariable limit
The value L that f(x, y) approaches as (x, y) approaches (a, b) along every possible path. In simple terms, no matter how you walk toward the point, you end up at the same number.
Continuity at a point
f is continuous at (a, b) if the limit as (x, y) approaches (a, b) equals f(a, b). Think of it as: the function has no jumps or holes at that point.
Path test
A method for investigating limits by substituting a specific path (y = ax, y = x², polar coordinates, etc.) into the function. Two paths giving different limits prove the limit does not exist. All paths giving the same limit do not, by themselves, prove it does.
Partial derivative
The derivative of f with respect to one variable while the others are held constant. f_x(a, b) is the slope of f in the x-direction at (a, b). Think of it as slicing the surface with a plane and measuring the slope of the resulting curve.
Tangent plane
The plane that best approximates the surface z = f(x, y) near a point. It is the two-variable analogue of the tangent line.
Linear approximation
Using the tangent plane to estimate f near a known point: f(x, y) ≈ f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b).
Gradient
∇f = ⟨f_x, f_y⟩ (or ⟨f_x, f_y, f_z⟩ in three variables). The gradient points in the direction of steepest increase and is perpendicular to level curves.
Directional derivative
The rate of change of f in the direction of a unit vector u: D_u f = ∇f · u. In simple terms, it tells you how fast f changes if you walk in a chosen direction.
Chain rule (multivariable)
If z = f(x, y) and x = x(t), y = y(t), then dz/dt = f_x · dx/dt + f_y · dy/dt. Think of it as: the total rate of change adds up the contributions from each variable.
Level surface / level set
The set of points where a function takes a constant value: T(x, y, z) = c defines a surface in R³. For T(x,y,z) = 2x + y + 3z + 27, the level surface T = 30 is the plane 2x + y + 3z = 3.
The piecewise function f(x, y) = x³y² / (x⁶ + y²) with f(0,0) = 0
This is the exam's workhorse example for understanding why path tests are tricky.
Along straight lines y = ax:
Substitute: f(x, ax) = x³(ax)² / (x⁶ + (ax)²) = a²x⁵ / (x⁶ + a²x²) = a²x³ / (x⁴ + a²).
As x → 0, this approaches 0 for every value of a.
So every straight-line path gives limit 0.
Along the curve y = x³:
Substitute: f(x, x³) = x³(x³)² / (x⁶ + (x³)²) = x⁹ / (x⁶ + x⁶) = x⁹ / (2x⁶) = x³/2.
As x → 0, this approaches 0 as well.
In this particular example, both families of paths give 0, and the function is in fact continuous at the origin (you can confirm with a squeeze argument). But the critical lesson is this:
Why path evidence alone is insufficient to prove a limit exists:
There are uncountably many paths into any point. Getting the same value on every straight line and on y = x² does not rule out a different value on some other exotic path. To prove a limit exists, you need a global argument: the squeeze theorem, conversion to polar coordinates with a bound that goes to zero regardless of angle, or the epsilon-delta definition.
To prove a limit does not exist, you only need two paths that disagree. That is sufficient.
Level surfaces
For T(x, y, z) = 2x + y + 3z + 27, the level surface T = 30 is found by setting 2x + y + 3z + 27 = 30, which gives 2x + y + 3z = 3. The correct multiple-choice answer is (A).
Tangent plane equation
For z = f(x, y) at the point (a, b, f(a, b)):
z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)
Worked example: f(2, 5) = 7, f_x(2, 5) = 3, f_y(2, 5) = -2
Tangent plane: z = 7 + 3(x - 2) + (-2)(y - 5) = 7 + 3x - 6 - 2y + 10 = 3x - 2y + 11.
Linear approximation of f(2.04, 4.97):
Δx = 0.04, Δy = -0.03.
f ≈ 7 + 3(0.04) + (-2)(-0.03) = 7 + 0.12 + 0.06 = 7.18.
The correct multiple-choice answer is (A) 7.18.
The chain-rule viewpoint:
The linear approximation is exactly what the chain rule produces if you think of (x, y) as moving along a short path from (2, 5) to (2.04, 4.97). The total change in f is approximately f_x·Δx + f_y·Δy, which is the differential df.
The gradient and the distance to a level plane
For T(x, y, z) = 2x + y + 3z + 27:
∇T = ⟨2, 1, 3⟩ (constant, since T is linear).
|∇T| = √(4 + 1 + 9) = √14.
The level plane T = 30 is 2x + y + 3z = 3.
Distance from (0, 1, 2) to this plane: |2(0) + 1(1) + 3(2) - 3| / √14 = |1 + 6 - 3| / √14 = 4/√14.
Directional derivative
With ∇f = ⟨3, -2⟩ at (2, 5) and unit direction u = ⟨3/5, 4/5⟩:
D_u f = ∇f · u = 3(3/5) + (-2)(4/5) = 9/5 - 8/5 = 1/5.
The rule
If z = f(x, y), x = x(t), y = y(t), then:
dz/dt = f_x · dx/dt + f_y · dy/dt
Worked example
Given T_x(2, 3) = 4, T_y(2, 3) = -3, x(t) = 1 + t, y(t) = 2 - 3t, and the position is (2, 3) at the relevant time:
dx/dt = 1, dy/dt = -3.
dT/dt = 4(1) + (-3)(-3) = 4 + 9 = 13.
The chain rule tells you that the temperature changes at rate 13 per unit time as you move along the path through (2, 3).
Tangent plane: z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)
Linear approximation: Δf ≈ f_x Δx + f_y Δy
Gradient: ∇f = ⟨f_x, f_y⟩ (2D) or ⟨f_x, f_y, f_z⟩ (3D)
Directional derivative: D_u f = ∇f · u (where u is a unit vector)
Chain rule: dz/dt = f_x · dx/dt + f_y · dy/dt
Distance from point to plane: d = |ax₀ + by₀ + cz₀ - d| / √(a² + b² + c²)
Level surface: T(x,y,z) = c defines the surface where T is constant at value c.
Linear approximation is how engineers estimate small changes without re-running expensive calculations: if you know the temperature at one point and its partial derivatives, you can estimate the temperature a short distance away. The gradient is used in machine learning (gradient descent) and in physics (heat flows in the direction opposite to the temperature gradient). Directional derivatives tell you the rate of change in any direction, which matters in terrain analysis, fluid flow, and optimisation.
Students often believe that checking every straight-line path is enough to prove a limit exists. It is not. You must also consider curved paths (y = x², y = x³, etc.) or use a global method.
Mixing up ∇f (a vector) with D_u f (a scalar). The gradient is a vector; the directional derivative is a number.
Forgetting that u must be a unit vector in D_u f = ∇f · u. If the given direction is not unit length, normalise it first.
Confusing the tangent plane equation with the equation of the level surface. The tangent plane approximates z = f(x,y) near a point; a level surface is where a function of three variables equals a constant.
⚠️ "Explain why the path evidence is sufficient or insufficient" is a near-certain exam question. Know the distinction: two disagreeing paths prove non-existence; agreeing paths prove nothing on their own.
⚠️ The tangent-plane and linear-approximation computation (plugging in Δx, Δy) is quick marks. Do not lose them to sign errors.
⚠️ The level surface question T = 30 for T = 2x + y + 3z + 27 is a one-step subtraction: 2x + y + 3z = 3. The exam answer is (A).
⚠️ The chain rule question will often give you partial derivatives at a point and a parametric path. Plug and add.
True or false: If f(x, y) approaches 0 along every line y = ax, the limit at the origin is 0. (Answer: False, you also need to check non-linear paths and/or use a global argument.)
The tangent plane to z = f(x,y) at (2, 5, 7) with f_x = 3 and f_y = -2 is z = ___. (Answer: 3x - 2y + 11)
Fill in the blank: D_u f = ___ · ___. (Answer: ∇f · u)
True or false: The gradient vector is tangent to level curves. (Answer: False, it is perpendicular to them.)
For T = 2x + y + 3z + 27, the level surface T = 30 is ___. (Answer: 2x + y + 3z = 3)
Q: For f(x,y) = x³y²/(x⁶ + y²) when (x,y) ≠ (0,0) and f(0,0) = 0, compute the limit along y = ax and along y = x³. Is f continuous at the origin?
A: Along y = ax: f = a²x³/(x⁴ + a²) → 0. Along y = x³: f = x³/2 → 0. Both give 0, consistent with continuity. However, to prove continuity you need a global argument (e.g. squeeze theorem), since path agreement alone is insufficient.
Q: Find the tangent plane to z = f(x,y) at (2, 5, 7) given f_x(2,5) = 3 and f_y(2,5) = -2. Use it to approximate f(2.04, 4.97).
A: Tangent plane: z = 7 + 3(x - 2) - 2(y - 5). At (2.04, 4.97): z ≈ 7 + 3(0.04) - 2(-0.03) = 7 + 0.12 + 0.06 = 7.18.
Q: State the multivariable chain rule and compute dT/dt at (2, 3) given T_x = 4, T_y = -3, x(t) = 1+t, y(t) = 2-3t.
A: dT/dt = T_x · dx/dt + T_y · dy/dt = 4(1) + (-3)(-3) = 4 + 9 = 13.
Q: What is the level surface T = 30 for T(x,y,z) = 2x + y + 3z + 27?
A: 2x + y + 3z = 3.
The tangent plane reappears in constrained optimisation (Lagrange multipliers use the gradient). The chain rule underpins related-rates problems in higher dimensions and the derivation of the heat equation. Directional derivatives and the gradient connect directly to line integrals and potential functions later in the course.
multivariable limit, two-variable limit, path test, continuity in two variables, piecewise function limit, tangent plane, linear approximation, linearisation, differential df, partial derivative, gradient vector, nabla, directional derivative, chain rule multivariable, level set, level surface, level curve, MATH 215, Calculus III, University of Michigan