This deck covers functions of two and three variables: domain, range, graphs as surfaces, level curves and level surfaces, and the limits story. It targets three misconceptions: that two agreeing paths prove a limit exists, that a boundary point can be substituted freely, and that a 0/0 form automatically means there is no limit.
Subject: Calculus III · 112 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
This week we leave single-variable calculus behind. Our inputs become points in the plane (or in space), and our outputs stay single numbers.
1. Find the natural domain and range of a function of two or three variables.
2. Read and sketch level curves (contour maps) and describe level surfaces.
3. Decide whether a limit exists by testing paths, and prove it does not exist with two disagreeing paths.
4. Confirm a limit exists using the squeeze idea and polar coordinates, and test continuity.
Warm-up
Discussion prompt
Before we open Week 5 - Functions of Several Variables: without looking back, what was the main idea of Week 4 - Unit Tangent, Arc Length & Curvature, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers the unit tangent and principal unit normal vectors, the arc length of a space curve as the integral of speed, and the arc-length parameter. It then gives three curvature formulas, the radius of curvature, the osculating circle, and the split of acceleration into its tangential and normal components. It targets the classic errors: skipping normalization before finding N, integrating the wrong quantity for arc length, and reaching for the wrong curvature formula.
Concept
A function of two variables takes an ordered pair of inputs and returns exactly one number.
\[ z = f(x,y) \]
Here x and y are the two independent inputs and z is the single dependent output.
function of two variables — A rule that assigns to each ordered pair (x, y) in a set D exactly one real number f(x, y). D is the domain; the collection of outputs is the range.
Counterexample
Discussion prompt
A function of two variables takes an ordered pair of inputs and returns exactly one number.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Intuition
Stand on a flat floor. Every spot on the floor is a point with two coordinates.
A function of two variables assigns each spot a height above (or below) the floor, like the temperature at each point of a room or the elevation at each point on a map.
One number per spot. The whole landscape of heights is what we will learn to picture.
Analogy
Discussion prompt
Explain Picture a height over the floor by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Stand on a flat floor. Every spot on the floor is a point with two coordinates.
Ranking
Put in order
Put the moves of Evaluating a function at a point into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Replace x with 2 and y with 3 everywhere.
Worked example
Evaluate the function below at the point where the first input is 2 and the second is 3.
\[ f(x,y) = x^2 + xy - y \]
Substitute the inputs
Why: Replace x with 2 and y with 3 everywhere.
\[ f(2,3) = (2)^2 + (2)(3) - 3 \]
Simplify term by term
Why: Square first, then the product, then combine.
\[ = 4 + 6 - 3 = 7 \]
Verify by re-adding the pieces
Why: Recompute independently: 4 plus 6 is 10, minus 3 is 7. The output at (2, 3) is 7.
\[ f(2,3) = 7 \]
Picture it
Animation
Shows: Each line of the worked example "Evaluating a function at a point", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Recompute independently: 4 plus 6 is 10, minus 3 is 7. The output at (2, 3) is 7.
Concept
Nothing stops us at two inputs. A function of three variables takes a point in space and returns a number.
\[ w = f(x,y,z) \]
We cannot draw its graph (that would need four dimensions), but it still has a domain, a range, limits, and continuity, exactly like the two-variable case.
Explain it
Discussion prompt
Explain Three or more variables to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Nothing stops us at two inputs. A function of three variables takes a point in space and returns a number.
Concept
When only a formula is given, the natural domain is every input for which the formula produces a real number.
natural domain — The largest set of input points for which the given formula is defined as a real number.
Finding it means asking: which points would break the formula? Those points, and only those, are thrown out.
Intuition
Almost every domain question comes down to three familiar troublemakers.
1. An even root cannot accept a negative inside.
2. A fraction cannot have a zero on the bottom.
3. A logarithm cannot accept a value that is zero or negative.
Track down where each of these would happen, and the leftover points are your domain.
Concept
A square root (or any even root) needs the quantity underneath to be zero or positive.
\[ \sqrt{g(x,y)} \text{ requires } g(x,y) \ge 0 \]
This usually carves the plane into a region bounded by a curve, keeping the side where the inside is nonnegative.
Concept
A fraction is undefined exactly where its denominator is zero.
\[ \frac{1}{h(x,y)} \text{ requires } h(x,y) \ne 0 \]
So we delete the curve (or points) where the bottom equals zero, and keep everything else.
Concept
A logarithm only accepts strictly positive inputs.
\[ \ln(g(x,y)) \text{ requires } g(x,y) > 0 \]
Note the strict inequality: zero is not allowed for a logarithm, unlike the even root where zero is fine.
Step zero
Discussion prompt
Domain of a square-root function — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Demand the inside be nonnegative
Answer:
Worked example
Find the natural domain of this function.
\[ f(x,y) = \sqrt{9 - x^2 - y^2} \]
Demand the inside be nonnegative
Why: An even root requires its radicand to be zero or greater.
\[ 9 - x^2 - y^2 \ge 0 \]
Rearrange into a familiar shape
Why: Move the squares to the other side to recognize a circle.
\[ x^2 + y^2 \le 9 \]
Describe the region
Why: This is the closed disk of radius 3 centered at the origin, boundary included.
Figure (svg): A filled disk of radius 3 centered at the origin representing the domain
Verify a boundary and an outside point
Why: At (3, 0): 9 minus 9 minus 0 is 0, and the root of 0 is defined, so the boundary belongs. At (3, 1): 9 minus 9 minus 1 is negative, so it is excluded. The description holds.
\[ \{(x,y) : x^2 + y^2 \le 9\} \]
Notation
Annotate
From Domain of a square-root function — read this one piece at a time. What is each part doing?
On: \( f(x,y) = \sqrt{9 - x^2 - y^2} \)
Estimation
Predict first
Find the natural domain of this rational function.
Commit before you compute: what does Domain with a denominator come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with a point on and off the parabola
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At (2, 4): the bottom is 4 minus 4, which is 0, so it is excluded.
Worked example
Find the natural domain of this rational function.
\[ f(x,y) = \frac{x}{x^2 - y} \]
Forbid a zero denominator
Why: The fraction is undefined wherever the bottom vanishes.
\[ x^2 - y \ne 0 \]
Identify the excluded curve
Why: Solve the equation the denominator would satisfy to see what to remove.
\[ y \ne x^2 \]
Verify with a point on and off the parabola
Why: At (2, 4): the bottom is 4 minus 4, which is 0, so it is excluded. At (2, 3): the bottom is 1, which is fine. The domain is the whole plane except the parabola y equals x squared.
\[ \{(x,y) : y \ne x^2\} \]
Picture it
Animation
Shows: Each line of the worked example "Domain with a denominator", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At (2, 4): the bottom is 4 minus 4, which is 0, so it is excluded. At (2, 3): the bottom is 1, which is fine. The domain is the whole plane except the parabola y equals x squared.
Fill the middle
Fill in the blanks
From Domain with a logarithm — finish the line. Write what belongs on the right of the equals sign before you look.
f(x,y) = \ln(x - y)
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. A logarithm accepts only strictly positive inputs.
Worked example
Find the natural domain of this function.
\[ f(x,y) = \ln(x - y) \]
Require a positive argument
Why: A logarithm accepts only strictly positive inputs.
\[ x - y > 0 \]
Read it as a half-plane
Why: Rewrite to see which side of the line is kept.
\[ x > y \]
Verify a point on each side
Why: At (2, 1): 2 minus 1 is 1, positive, so it is in. At (1, 2): 1 minus 2 is negative, so it is out. The domain is the open half-plane where x exceeds y; the line itself is excluded.
\[ \{(x,y) : x > y\} \]
Picture it
Animation
Shows: Each line of the worked example "Domain with a logarithm", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At (2, 1): 2 minus 1 is 1, positive, so it is in. At (1, 2): 1 minus 2 is negative, so it is out. The domain is the open half-plane where x exceeds y; the line itself is excluded.
Concept
The range is the set of all output values the function actually produces as the inputs sweep over the whole domain.
range — The set of all values f(x, y) takes as (x, y) runs over the domain.
To find it, ask how small and how large the output can get, and whether every value in between is reached.
Missing information
Discussion prompt
Find the range of the function from before, whose domain is the disk of radius 3.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
On the disk, the quantity x squared plus y squared runs from 0 at the center to 9 at the boundary.
Worked example
Find the range of the function from before, whose domain is the disk of radius 3.
\[ f(x,y) = \sqrt{9 - x^2 - y^2} \]
Bound the inside of the root
Why: On the disk, the quantity x squared plus y squared runs from 0 at the center to 9 at the boundary.
\[ 0 \le 9 - x^2 - y^2 \le 9 \]
Apply the square root
Why: The root is increasing, so the smallest and largest outputs come from the smallest and largest insides.
\[ 0 \le \sqrt{9 - x^2 - y^2} \le 3 \]
Verify the endpoints are hit
Why: At the origin the output is the root of 9, which is 3; on the boundary circle the output is the root of 0, which is 0. Every value between is reached, so the range is the closed interval from 0 to 3.
\[ \text{range} = [0, 3] \]
Picture it
Animation
Shows: Each line of the worked example "Range of the square-root function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the origin the output is the root of 9, which is 3; on the boundary circle the output is the root of 0, which is 0. Every value between is reached, so the range is the closed interval from 0 to 3.
Pattern
1. Scan the formula for the three troublemakers
Why: Even roots, denominators, and logarithms are where a real-number output can fail.
2. Write the condition each one forces
Why: Root: inside at least zero. Denominator: not zero. Logarithm: strictly positive.
3. Combine the conditions
Why: All of them must hold at once, so intersect the regions they describe.
4. Describe the region in words and a sketch
Why: Name it: a disk, a half-plane, the plane minus a curve. Then test a point to confirm.
Elimination
Eliminate the wrong options
Which of these points is NOT in the domain?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: D
Why: The domain requires x squared plus y squared to be at most 9. At (3, 3) that sum is 9 plus 9, which is 18, far above 9, so the inside of the root is negative and the point is excluded.
Check
Use the disk domain you just found for the square-root function.
\[ f(x,y) = \sqrt{9 - x^2 - y^2} \]
Check your understanding
Which of these points is NOT in the domain?
Answer: D
Why: The domain requires x squared plus y squared to be at most 9. At (3, 3) that sum is 9 plus 9, which is 18, far above 9, so the inside of the root is negative and the point is excluded.
Trap
Careless move: evaluate the square-root function at a point where the inside is negative, as if the formula always works.
\[ f(x,y) = \sqrt{9 - x^2 - y^2}, \quad \text{try } (3,3) \]
Wrong: charge ahead and take a root of a negative number.
\[ \sqrt{9 - 9 - 9} = \sqrt{-9} \;? \]
This is not a real number, so the point was never a legal input in the first place.
Right: check membership in the domain before evaluating or before taking a limit there.
\[ x^2 + y^2 \le 9 \;\Rightarrow\; 18 \le 9 \text{ is false} \]
Since the point sits outside the disk, the function has no value there. Also beware boundary points: at a boundary point you can only approach through the domain, so a limit uses just the inside of the region.
\[ (3,3) \notin \{(x,y): x^2+y^2 \le 9\} \]
Concept
Draw the two input axes as a floor and stack the output up as height. The set of all points that satisfy the equation below forms a surface in space.
\[ z = f(x,y) \]
graph of f — The set of all points (x, y, z) in space with z equal to f(x, y), where (x, y) lies in the domain. It is a surface floating over the domain.
Definition probe
Sort into buckets
Every line below is part of the definition of function of two variables or of graph of f — one or the other, never both. Put each where it belongs.
Intuition
Think of the domain as a patch of floor and the graph as a bedsheet draped above it, dipping and rising.
Directly above each floor point, the sheet sits at exactly one height. That is what makes it a function: one height per spot, never two.
Concept
Surfaces are hard to draw freehand. A trick is to slice.
trace — The curve you get by intersecting the surface with a plane, for example by holding one variable fixed.
Hold the first input at a constant and you get a curve in the remaining input and the output; do the same for the second input. A few such slices reveal the shape.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of function of two variables, natural domain, range, graph of f, trace as Week 5 - Functions of Several Variables uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
There is a second slicing idea, and it is the star of this section: slice at a constant height.
\[ f(x,y) = c \]
level curve — The set of domain points where f(x, y) equals a fixed constant c. Drawing several, one per height, gives a contour map.
Intuition
A topographic map draws a loop for every fixed elevation. Walk along one loop and your height never changes.
That loop is exactly a level curve: the shadow, cast straight down onto the floor, of a horizontal slice through the surface.
Concept
Contours drawn at equally spaced heights carry extra information in how far apart they sit.
Where the curves crowd together, the surface climbs steeply. Where they spread far apart, the surface is nearly flat.
Step zero
Discussion prompt
Level curves: circles — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set the function equal to a constant
Answer:
Worked example
Describe the level curves of this function.
\[ f(x,y) = x^2 + y^2 \]
Set the function equal to a constant
Why: A level curve is where the output holds fixed at some value c.
\[ x^2 + y^2 = c \]
Interpret by the sign of c
Why: For c positive this is a circle; for c equal to 0 it is the single origin; for c negative there are no points, since a sum of squares cannot be negative.
\[ c > 0:\ \text{circle of radius } \sqrt{c} \]
Note the spacing
Why: Equal steps in c give circles whose radii grow like the square root, so they crowd together as you move outward, meaning the bowl gets steeper.
Figure (svg): Concentric circles centered at the origin representing level curves
Verify one curve
Why: For c equal to 4 the radius should be 2. The point (2, 0) gives 4 plus 0 equals 4, and it sits distance 2 from the origin. The description checks out: the level curves are concentric circles.
\[ c = 4:\ x^2 + y^2 = 4,\ \text{radius } 2 \]
Notation
Annotate
From Level curves: circles — read this one piece at a time. What is each part doing?
On: \( c = 4:\ x^2 + y^2 = 4,\ \text{radius } 2 \)
Estimation
Predict first
Describe the level curves of this function.
Commit before you compute: what does Level curves: parabolas come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify one curve
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. For c equal to 0 the curve is y equals x squared.
Worked example
Describe the level curves of this function.
\[ f(x,y) = y - x^2 \]
Set equal to a constant
Why: Fix the output at c to find one level curve.
\[ y - x^2 = c \]
Solve for the output variable
Why: Isolate y to recognize the shape.
\[ y = x^2 + c \]
Describe the family
Why: Each level curve is an upward parabola, the base parabola shifted up or down by c.
\[ \text{parabolas } y = x^2 + c \]
Verify one curve
Why: For c equal to 0 the curve is y equals x squared. The point (1, 1) gives 1 minus 1 equals 0, matching c equals 0. The family is vertically shifted parabolas.
\[ c=0:\ (1,1)\ \text{gives}\ 1-1=0 \]
Picture it
Animation
Shows: Each line of the worked example "Level curves: parabolas", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For c equal to 0 the curve is y equals x squared. The point (1, 1) gives 1 minus 1 equals 0, matching c equals 0. The family is vertically shifted parabolas.
Hypothesis
Predict first
Level curves: hyperbolas is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Set equal to a nonzero constant
Why: Fix the product at c to trace one curve.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Describe the level curves of this function.
\[ f(x,y) = xy \]
Set equal to a nonzero constant
Why: Fix the product at c to trace one curve.
\[ xy = c \]
Solve for one variable
Why: For c not zero, dividing shows the reciprocal shape.
\[ y = \frac{c}{x} \]
Describe the family
Why: Each nonzero level is a hyperbola with the axes as asymptotes; the zero level is the pair of coordinate axes themselves.
\[ c \ne 0:\ \text{hyperbolas } y = c/x \]
Verify a point
Why: For c equal to 1 the point (2, 0.5) gives the product 1, and it lies on y equals 1 over x. The nonzero level curves are hyperbolas.
\[ c=1:\ (2, 0.5)\ \text{gives}\ 2\cdot 0.5 = 1 \]
Picture it
Animation
Shows: Each line of the worked example "Level curves: hyperbolas", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For c equal to 1 the point (2, 0.5) gives the product 1, and it lies on y equals 1 over x. The nonzero level curves are hyperbolas.
Concept
A function of three variables cannot be graphed, but the same fix-the-output idea still works.
\[ f(x,y,z) = c \]
level surface — The set of space points where a three-variable function equals a fixed constant c. It is a surface, the three-variable analogue of a level curve.
Ranking
Put in order
Put the moves of Level surfaces: spheres into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Hold the output at c to find one level surface.
Worked example
Describe the level surfaces of this function.
\[ f(x,y,z) = x^2 + y^2 + z^2 \]
Set equal to a positive constant
Why: Hold the output at c to find one level surface.
\[ x^2 + y^2 + z^2 = c \]
Recognize the shape
Why: This is the standard equation of a sphere centered at the origin whose radius is the square root of c.
\[ c > 0:\ \text{sphere of radius } \sqrt{c} \]
Verify one surface
Why: For c equal to 1 the radius should be 1. The point (1, 0, 0) gives 1 plus 0 plus 0 equals 1 and sits distance 1 from the origin. The level surfaces are concentric spheres.
\[ c=1:\ (1,0,0)\ \text{gives}\ 1 \]
Picture it
Animation
Shows: Each line of the worked example "Level surfaces: spheres", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For c equal to 1 the radius should be 1. The point (1, 0, 0) gives 1 plus 0 plus 0 equals 1 and sits distance 1 from the origin. The level surfaces are concentric spheres.
Prediction
Predict first
The level curves f(x, y) = c for c greater than 0 are:
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Circles centered at the origin
Why: Setting x squared plus y squared equal to a positive constant c gives the equation of a circle centered at the origin with radius equal to the square root of c, so the contours are concentric circles.
Check
Consider the level curves of the following function.
\[ f(x,y) = x^2 + y^2 \]
Check your understanding
The level curves f(x, y) = c for c greater than 0 are:
Answer: A
Why: Setting x squared plus y squared equal to a positive constant c gives the equation of a circle centered at the origin with radius equal to the square root of c, so the contours are concentric circles.
Concept
A limit asks a simple question: as the input point slides toward a target point, does the output settle on a single number?
\[ \lim_{(x,y) \to (a,b)} f(x,y) = L \]
limit of f at (a, b) — The single value L that f(x, y) approaches as (x, y) gets arbitrarily close to (a, b), no matter how the point approaches.
Intuition
In one variable, you could only slide in from the left or from the right. Two numbers to match.
In the plane, the target point is surrounded. You can walk in along a straight line from any angle, or along a curve that spirals, bends, or wiggles.
For the limit to exist, all of these infinitely many approaches must land on the same output value.
Concept
This is the single most important idea of the section.
If the limit exists and equals L, then the output must approach L along every possible path to the point.
Turn that around and it becomes a powerful tool: if you can find even two paths that give different values, the limit cannot exist.
Intuition
The one-variable rule was: left-hand limit must equal right-hand limit. Just two things to compare.
The two-variable rule is the same spirit, but now there are infinitely many routes in. Agreement of every route is a much stronger demand, and that is why these limits are subtle.
Concept
A function is continuous at a point when three things all line up: the value exists, the limit exists, and they are equal.
\[ \lim_{(x,y)\to(a,b)} f(x,y) = f(a,b) \]
continuous at (a, b) — f is continuous at (a, b) if f(a, b) is defined, the limit there exists, and the limit equals f(a, b). No holes, no jumps.
Concept
Polynomials in two variables are continuous everywhere. Rational functions, sums, products, and compositions of continuous functions are continuous wherever they are defined.
When a function is continuous at the target point, the limit is found by simple substitution: just plug the point in.
\[ \lim_{(x,y)\to(a,b)} f(x,y) = f(a,b) \]
Worked example
Evaluate the limit of a polynomial as the point approaches (1, 2).
\[ \lim_{(x,y)\to(1,2)} \left( x^2 + 3xy \right) \]
Note the function is a polynomial
Why: Polynomials in two variables are continuous everywhere, so the limit is the value at the point.
Substitute the point
Why: Replace x with 1 and y with 2.
\[ (1)^2 + 3(1)(2) \]
Simplify
Why: Square, multiply, then add.
\[ 1 + 6 = 7 \]
Verify by continuity
Why: Because the polynomial is continuous at (1, 2), substitution is valid and no path test is needed. The limit is 7.
\[ \lim_{(x,y)\to(1,2)} (x^2 + 3xy) = 7 \]
Picture it
Animation
Shows: Each line of the worked example "A limit by substitution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Because the polynomial is continuous at (1, 2), substitution is valid and no path test is needed. The limit is 7.
Missing information
Discussion prompt
Evaluate this rational limit as the point approaches (1, 2).
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
A rational function is continuous wherever the denominator is nonzero. Here y minus x is 2 minus 1, which is 1, not zero.
Worked example
Evaluate this rational limit as the point approaches (1, 2).
\[ \lim_{(x,y)\to(1,2)} \frac{x+y}{y-x} \]
Check the denominator at the point
Why: A rational function is continuous wherever the denominator is nonzero. Here y minus x is 2 minus 1, which is 1, not zero.
\[ y - x = 2 - 1 = 1 \ne 0 \]
Substitute the point
Why: Since the denominator is nonzero, the function is continuous there, so plug in directly.
\[ \frac{1+2}{2-1} = \frac{3}{1} \]
Verify the value
Why: The denominator was safely nonzero, so substitution is justified. The limit is 3.
\[ \lim_{(x,y)\to(1,2)} \frac{x+y}{y-x} = 3 \]
Picture it
Animation
Shows: Each line of the worked example "A rational limit by substitution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A rational function is continuous wherever the denominator is nonzero. Here y minus x is 2 minus 1, which is 1, not zero.
Trap
Tempting shortcut: plug the point in, see the indeterminate form, and declare defeat.
\[ \lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x - y} \]
Wrong: at the origin the numerator and denominator both hit 0, so the student writes zero over zero and concludes the limit does not exist.
\[ \frac{0-0}{0-0} = \frac{0}{0}\ \Rightarrow\ \text{DNE?} \]
Right: the indeterminate form only says substitution failed, not that the limit is gone. Simplify first.
\[ \frac{x^2 - y^2}{x - y} = \frac{(x-y)(x+y)}{x-y} = x + y \]
For every point off the line where x equals y, the function equals x plus y, which approaches 0. The limit exists and is 0.
\[ \lim_{(x,y)\to(0,0)} (x+y) = 0 \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Right: the indeterminate form only says substitution failed, not that the limit is gone. Simplify first.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
When substitution gives an indeterminate form and simplifying does not clear it, suspect the limit may not exist and reach for the two-path test.
Choose two convenient paths to the point. Compute the limit of the function along each. If the two answers differ, the limit does not exist.
two-path test — If f approaches different values along two different paths to (a, b), then the limit at (a, b) does not exist.
Intuition
You do not need to check every path to break a limit. Two honest paths that disagree already contradict the all-paths principle.
The natural first paths to try are the simplest: along the x-axis, along the y-axis, and along lines through the point.
Step zero
Discussion prompt
Showing a limit does not exist — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Approach along the x-axis
Answer:
Worked example
Investigate this limit at the origin.
\[ \lim_{(x,y)\to(0,0)} \frac{xy}{x^2 + y^2} \]
Approach along the x-axis
Why: Set y equal to 0. Then the numerator is 0 for every x, so the function is 0 all along this path.
\[ y=0:\ \frac{x\cdot 0}{x^2 + 0} = 0 \to 0 \]
Approach along the line y equals x
Why: Substitute y equal to x. The numerator is x times x and the denominator is x squared plus x squared.
\[ y=x:\ \frac{x\cdot x}{x^2 + x^2} = \frac{x^2}{2x^2} = \frac{1}{2} \to \frac{1}{2} \]
Compare the two results
Why: One path gives 0 and the other gives one half. Different values along different paths violate the all-paths principle.
\[ 0 \ne \tfrac{1}{2} \]
Verify the contradiction
Why: Both computations are exact, and they disagree, so by the two-path test the limit does not exist.
\[ \lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2}\ \text{does not exist} \]
Picture it
Animation
Shows: Each line of the worked example "Showing a limit does not exist", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both computations are exact, and they disagree, so by the two-path test the limit does not exist.
Worked example
Investigate this limit at the origin.
\[ \lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x^2 + y^2} \]
Approach along the x-axis
Why: Set y equal to 0. The expression becomes x squared over x squared, which is 1.
\[ y=0:\ \frac{x^2}{x^2} = 1 \to 1 \]
Approach along the y-axis
Why: Set x equal to 0. The expression becomes negative y squared over y squared, which is negative 1.
\[ x=0:\ \frac{-y^2}{y^2} = -1 \to -1 \]
Verify the disagreement
Why: The two axis paths give 1 and negative 1, which differ, so the limit does not exist by the two-path test.
\[ 1 \ne -1 \Rightarrow \text{DNE} \]
Picture it
Animation
Shows: Each line of the worked example "Nonexistence along the axes", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two axis paths give 1 and negative 1, which differ, so the limit does not exist by the two-path test.
Trap
Dangerous reasoning: test a couple of straight lines, watch them agree, and declare the limit found.
\[ \lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^4 + y^2} \]
Along the x-axis the value is 0, and along any line through the origin the value also tends to 0.
\[ y = mx:\ \frac{m x^3}{x^4 + m^2 x^2} = \frac{mx}{x^2 + m^2} \to 0 \]
Wrong conclusion: since every line gives 0, the limit is 0.
Right: agreement on many paths proves nothing. You must suspect a cleverer path. Try the parabola where y equals x squared.
\[ y = x^2:\ \frac{x^2\cdot x^2}{x^4 + (x^2)^2} = \frac{x^4}{2x^4} = \frac{1}{2} \]
The parabola gives one half while the lines give 0. Since two paths disagree, the limit does not exist. Agreeing paths can never establish existence; disagreeing paths do establish nonexistence.
\[ 0 \ne \tfrac{1}{2} \Rightarrow \text{DNE} \]
Concept
Paths can only ever disprove a limit. To actually prove one exists, you must control the function for all approaches at once.
Two reliable tools do this: the squeeze idea, which traps the function between bounds that both shrink to the target, and a switch to polar coordinates, which measures distance to the point with a single variable.
Intuition
Imagine the output caught between a floor and a ceiling that both slide toward the same number. The output has nowhere to go but that number.
\[ 0 \le |f(x,y)| \le g(x,y), \quad g \to 0 \;\Rightarrow\; f \to 0 \]
The skill is finding a simple bound g that clearly shrinks to zero no matter which direction you came from.
Concept
Near the origin, polar coordinates measure exactly what matters: the distance from the origin is one variable, and the direction is the other.
\[ x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2 \]
Approaching the origin means the distance goes to zero. If the rewritten function goes to a single number as the distance shrinks, with no leftover dependence on the direction, the limit exists and equals that number.
Estimation
Predict first
Show this limit exists and find its value.
Commit before you compute: what does A limit that exists, via polar come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by squeezing as r goes to 0
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The bound r goes to 0 with no dependence on the angle, so the function is squeezed to 0.
Worked example
Show this limit exists and find its value.
\[ \lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^2 + y^2} \]
Convert to polar
Why: Substitute the polar forms. The numerator becomes r squared times cosine squared, times r sine; the denominator is r squared.
\[ \frac{(r\cos\theta)^2 (r\sin\theta)}{r^2} = \frac{r^3 \cos^2\theta \sin\theta}{r^2} \]
Simplify
Why: Cancel r squared, leaving one factor of r times bounded trig terms.
\[ = r\cos^2\theta \sin\theta \]
Bound it
Why: Cosine squared times sine is at most 1 in size, so the whole thing is at most r in size, regardless of direction.
\[ \left| r\cos^2\theta \sin\theta \right| \le r \]
Verify by squeezing as r goes to 0
Why: The bound r goes to 0 with no dependence on the angle, so the function is squeezed to 0. The limit exists and equals 0.
\[ \lim_{r\to 0} r\cos^2\theta\sin\theta = 0 \]
Picture it
Animation
Shows: Each line of the worked example "A limit that exists, via polar", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The bound r goes to 0 with no dependence on the angle, so the function is squeezed to 0. The limit exists and equals 0.
Fill the middle
Fill in the blanks
From Another polar limit — finish the line. Write what belongs on the right of the equals sign before you look.
\fracr\cos^3\theta___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The numerator is r cubed cosine cubed; the denominator is r squared.
Worked example
Evaluate this limit at the origin.
\[ \lim_{(x,y)\to(0,0)} \frac{x^3}{x^2 + y^2} \]
Convert to polar
Why: The numerator is r cubed cosine cubed; the denominator is r squared.
\[ \frac{r^3 \cos^3\theta}{r^2} = r\cos^3\theta \]
Bound the result
Why: Cosine cubed is at most 1 in size, so the expression is at most r in size for every angle.
\[ \left| r\cos^3\theta \right| \le r \]
Verify by letting r go to 0
Why: The bound r shrinks to 0 independent of the direction, so by squeezing the limit exists and equals 0.
\[ \lim_{(x,y)\to(0,0)} \frac{x^3}{x^2+y^2} = 0 \]
Picture it
Animation
Shows: Each line of the worked example "Another polar limit", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The bound r shrinks to 0 independent of the direction, so by squeezing the limit exists and equals 0.
Worked example
Evaluate this limit at the origin.
\[ \lim_{(x,y)\to(0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} \]
Name the combination
Why: Both the numerator and denominator depend only on the sum of squares, so let that sum be a single quantity u.
\[ u = x^2 + y^2 \ge 0 \]
Rewrite the limit in u
Why: As the point approaches the origin, u approaches 0 from above, turning this into a familiar one-variable limit.
\[ \lim_{u\to 0^+} \frac{\sin u}{u} \]
Apply the known limit
Why: The classic single-variable limit of sine of u over u is 1.
\[ \frac{\sin u}{u} \to 1 \]
Verify the value
Why: The substitution is valid because u genuinely tends to 0, and the reduced limit equals 1. So the two-variable limit is 1.
\[ \lim_{(x,y)\to(0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} = 1 \]
Picture it
Animation
Shows: Each line of the worked example "A limit via a substitution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The substitution is valid because u genuinely tends to 0, and the reduced limit equals 1. So the two-variable limit is 1.
Concept
You do not always need polar. Sometimes a single inequality between the parts does the job.
A useful fact: each squared variable is no larger than the sum of the squares, so a ratio of that form is at most 1 in size.
\[ \frac{y^2}{x^2 + y^2} \le 1 \]
Explain it
Discussion prompt
Explain The direct squeeze with an inequality to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
You do not always need polar. Sometimes a single inequality between the parts does the job.
Step zero
Discussion prompt
Squeeze with a plain bound — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split off the bounded factor
Answer:
Worked example
Evaluate this limit at the origin using an inequality.
\[ \lim_{(x,y)\to(0,0)} \frac{5xy^2}{x^2 + y^2} \]
Split off the bounded factor
Why: Write the expression as 5 times x times the ratio of y squared to the sum of squares.
\[ \left| \frac{5xy^2}{x^2+y^2} \right| = 5|x|\cdot \frac{y^2}{x^2+y^2} \]
Bound the ratio by 1
Why: Since y squared is at most the sum of squares, that ratio is at most 1, leaving 5 times the size of x.
\[ \le 5|x| \]
Verify by squeezing
Why: As the point approaches the origin, the size of x goes to 0, so 5 times it does too. The function is squeezed to 0, and the limit is 0.
\[ 5|x| \to 0 \;\Rightarrow\; \lim = 0 \]
Picture it
Animation
Shows: Each line of the worked example "Squeeze with a plain bound", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: As the point approaches the origin, the size of x goes to 0, so 5 times it does too. The function is squeezed to 0, and the limit is 0.
Pattern
1. Try direct substitution
Why: If the function is continuous at the point (a polynomial, or a rational with nonzero denominator there), just plug in and you are done.
2. If you get an indeterminate form, simplify
Why: Factor and cancel. An indeterminate form does not mean the limit is gone; a hidden common factor may reveal it.
3. Suspect nonexistence: test two paths
Why: Try the axes and a line. If two paths disagree, the limit does not exist and you are finished.
4. If paths keep agreeing, try to prove existence
Why: Agreement is not proof. Switch to polar or find a shrinking bound to squeeze the function to a single value.
Real world
Discussion prompt
Outside this lesson: where does Week 5 - Functions of Several Variables actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe: evaluate a two-variable limit is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers functions of two and three variables: domain, range, graphs as surfaces, level curves and level surfaces, and the limits story. It targets three misconceptions: that two agreeing paths prove a limit exists, that a boundary point can be substituted freely, and that a 0/0 form automatically means there is no limit.
Check
Evaluate the limit of a polynomial as the point approaches (2, 3).
\[ \lim_{(x,y)\to(2,3)} \left( xy - x^2 \right) \]
Check your understanding
What is the value of the limit?
Answer: A
Why: The function is a polynomial, hence continuous, so substitute directly: x times y is 2 times 3, which is 6, and x squared is 4, giving 6 minus 4, which equals 2.
Prediction
Predict first
What can you conclude about the limit at the origin?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: It does not exist
Why: Two paths to the origin give different values, 0 and one half. The all-paths principle says the limit would have to agree on every path, so having two that disagree proves the limit does not exist.
Check
For the function below, along the line where y equals x the value is one half, and along the x-axis the value is 0.
\[ f(x,y) = \frac{xy}{x^2 + y^2} \]
Check your understanding
What can you conclude about the limit at the origin?
Answer: A
Why: Two paths to the origin give different values, 0 and one half. The all-paths principle says the limit would have to agree on every path, so having two that disagree proves the limit does not exist.
Check
Evaluate this limit at the origin.
\[ \lim_{(x,y)\to(0,0)} \frac{x^3}{x^2 + y^2} \]
Check your understanding
What is the value of the limit?
Answer: A
Why: In polar form the expression becomes r times cosine cubed of the angle, which is at most r in size for every direction. As r shrinks to 0 the function is squeezed to 0, so the limit is 0.
Concept
Continuity is preserved by the usual operations: sums, differences, products, quotients (where the denominator is nonzero), and compositions of continuous functions are continuous.
So the discontinuities of a built-up function live only where a denominator vanishes or where an inside piece leaves the allowed domain.
Analogy
Discussion prompt
Explain Building continuous functions by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Continuity is preserved by the usual operations: sums, differences, products, quotients (where the denominator is nonzero), and compositions of continuous functions are continuous.
Ranking
Put in order
Put the moves of Where is a function continuous? into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The numerator and denominator are polynomials, continuous everywhere, so continuity can only fail where the denominator is zero.
Worked example
Determine where this function is continuous.
\[ f(x,y) = \frac{x + y}{x^2 - y} \]
Identify the only threat
Why: The numerator and denominator are polynomials, continuous everywhere, so continuity can only fail where the denominator is zero.
\[ x^2 - y = 0 \]
Solve for the bad set
Why: Set the denominator to zero and solve to name the curve of discontinuity.
\[ y = x^2 \]
Verify with a point on and off the curve
Why: At (1, 1) the denominator is 1 minus 1, which is 0, so continuity fails there, on the parabola. At (1, 0) the denominator is 1, so the function is continuous there. The function is continuous everywhere except on the parabola y equals x squared.
\[ \text{continuous for } y \ne x^2 \]
Picture it
Animation
Shows: Each line of the worked example "Where is a function continuous?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At (1, 1) the denominator is 1 minus 1, which is 0, so continuity fails there, on the parabola. At (1, 0) the denominator is 1, so the function is continuous there. The function is continuous everywhere except on the parabola y equals x squared.
Trap
Overconfident move: convert to polar, notice the r cancels, and announce the limit is whatever is left.
\[ \lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x^2 + y^2} \]
Wrong: in polar the r cancels completely, and the student reports the leftover expression as if it were a fixed number.
\[ \frac{r^2\cos^2\theta - r^2\sin^2\theta}{r^2} = \cos 2\theta \]
Right: the result must be free of the angle to count as a limit. Here it still depends on the direction, so different approaches give different values.
\[ \theta = 0:\ \cos 0 = 1; \quad \theta = \tfrac{\pi}{2}:\ \cos\pi = -1 \]
Because the polar expression depends on the angle, the limit does not exist. Polar only proves existence when the direction drops out entirely.
\[ 1 \ne -1 \Rightarrow \text{DNE} \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
A quick decision guide keeps you from grinding on the wrong approach.
If plugging in gives a real number, the function is continuous there and that number is the limit. If it gives an indeterminate form, first try to simplify.
If simplifying fails, test paths to try to break it; if the paths stubbornly agree, switch to polar or a squeeze bound to prove the value.
Counterexample
Discussion prompt
A quick decision guide keeps you from grinding on the wrong approach.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
If plugging in gives a real number, the function is continuous there and that number is the limit. If it gives an indeterminate form, first try to simplify.
Elimination
Eliminate the wrong options
The function fails to be continuous exactly where:
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The numerator and denominator are polynomials, so continuity can fail only where the denominator is zero. Setting x squared minus y equal to zero gives y equals x squared, the parabola where the function is undefined.
Check
Consider the following rational function.
\[ f(x,y) = \frac{x + y}{x^2 - y} \]
Check your understanding
The function fails to be continuous exactly where:
Answer: A
Why: The numerator and denominator are polynomials, so continuity can fail only where the denominator is zero. Setting x squared minus y equal to zero gives y equals x squared, the parabola where the function is undefined.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: find the natural domain · Recipe: evaluate a two-variable limit · A function of two variables · Picture a height over the floor · Three or more variables. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can find the natural domain of a function of several variables by ruling out even roots of negatives, zero denominators, and nonpositive logarithm inputs, and you can read off its range.
You can picture a two-variable function as a surface and map it with level curves, and extend the idea to level surfaces in space.
For limits, the guiding law is the all-paths principle. Two disagreeing paths prove nonexistence; agreeing paths prove nothing. To confirm a limit, squeeze it or switch to polar and make sure the direction drops out.
And remember the three traps: never plug in outside the domain, an indeterminate form is not an automatic dead end, and matching paths are never a proof of existence.
| Situation | What to do |
|---|---|
| Continuous at the point | Substitute directly |
| Indeterminate form | Factor and simplify first |
| Suspect no limit | Find two disagreeing paths |
| Confirm a limit | Squeeze or use polar coordinates |
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