Source: MATH 163 Abridged Class Notes, Section 14.2
Tags: Squeeze Theorem, polar coordinates, multivariable limits, proving limit existence, Calculus III, MATH 163, bounding inequalities
Difficulty: Intermediate | Prerequisites: Path testing for multivariable limits (companion notes), single-variable Squeeze Theorem from Calculus I, polar coordinates from Pre-Calculus or Calculus II
Path testing can show a limit does not exist, but it cannot prove one does. When every path you try gives the same answer, you need a different tool. The Squeeze Theorem and polar coordinate substitution are the two standard methods for proving a multivariable limit exists. This is the other half of the Section 14.2 toolkit. You should be comfortable with path testing before working through this material.
The Squeeze Theorem lets you trap a function between two simpler functions that both approach the same value, confirming the limit. Polar coordinates convert (x, y) approaching (0, 0) into r approaching 0, which turns a two-variable problem into a single-variable one. If the result depends only on r (not theta), the limit exists. If theta survives, the limit does not exist.
Squeeze Theorem (Sandwich Theorem / Pinching Theorem)
If g(x, y) <= f(x, y) <= h(x, y) near (a, b), and both g and h approach the same limit L as (x, y) approaches (a, b), then f must also approach L.
Think of it as: if you can trap the function between a floor and a ceiling that both squeeze down to the same value, the function has no choice but to go there too.
Polar coordinates
A coordinate system where a point in the plane is described by its distance r from the origin and its angle theta from the positive x-axis. The conversion is x = r cos(theta), y = r sin(theta).
In simple terms, this means: instead of tracking two independent variables x and y, you track how far you are from the origin (r) and which direction you are facing (theta).
Bounding inequality
An inequality of the form |x| / sqrt(x^2 + y^2) <= 1 that arises from the fact that sqrt(x^2 + y^2) >= |x|. These inequalities are the building blocks for Squeeze Theorem arguments in 2D.
Think of it as: a way to control how big part of a fraction can get, relative to the whole expression.
Conjugate multiplication
Multiplying the numerator and denominator by the conjugate of a square-root expression to simplify the limit. Used when the denominator has the form sqrt(something) - 1.
In simple terms, this means: the algebraic trick of turning sqrt(a) - b into something that cancels nicely, so you can evaluate the limit directly.
The Squeeze Theorem works the same way in one variable and in two. The single-variable version is a good warm-up.
Consider the limit of x^2 cos(1/x) as x approaches 0. The cosine function is bounded: -1 <= cos(1/x) <= 1. Multiplying through by x^2 (which is non-negative) gives -x^2 <= x^2 cos(1/x) <= x^2. As x approaches 0, both -x^2 and x^2 approach 0, so by the Squeeze Theorem, the limit is 0.
The key insight: you do not need to know what cos(1/x) does as x approaches 0 (it oscillates wildly). You only need to know it is bounded, and that the rest of the expression goes to 0 fast enough to squash everything.
The same logic applies to multivariable limits. The strategy has three steps:
Find a bounding inequality. The most common one is |x| / sqrt(x^2 + y^2) <= 1, which follows from the fact that sqrt(x^2 + y^2) >= sqrt(x^2) = |x|.
Use that inequality to trap f(x, y) between two expressions that both approach the same limit.
Conclude by the Squeeze Theorem.
Evaluate the limit as (x, y) approaches (0, 0).
Since sqrt(x^2 + y^2) >= |x|, we have |x| / sqrt(x^2 + y^2) <= 1, which means -1 <= x / sqrt(x^2 + y^2) <= 1.
Case 1, y > 0: multiply the inequality by y to get -y <= xy / sqrt(x^2 + y^2) <= y. Both -y and y approach 0 as (x, y) approaches (0, 0), so the limit is 0 by the Squeeze Theorem.
Case 2, y < 0: the inequalities flip direction (because multiplying by a negative reverses the inequality), but the same reasoning gives the same conclusion: the limit is 0.
Case 3, y = 0: the function is 0 / sqrt(x^2) = 0, so the limit is trivially 0.
All cases agree. The limit is 0.
Polar coordinates offer a mechanical alternative to the Squeeze Theorem. The idea: approaching (0, 0) in Cartesian coordinates is the same as letting r approach 0 from the right in polar coordinates, with theta free to be anything.
Substitute x = r cos(theta), y = r sin(theta), and simplify. Two outcomes are possible:
If the result depends only on r (not theta) and approaches a finite value as r approaches 0, the limit exists and equals that value.
If theta survives in the expression after taking the limit as r approaches 0, the answer changes with the angle of approach, and the limit does not exist.
Substitute: (r cos theta)(r sin theta) / sqrt(r^2 cos^2 theta + r^2 sin^2 theta) = r^2 cos theta sin theta / sqrt(r^2) = r^2 cos theta sin theta / r = r cos theta sin theta.
As r approaches 0, this approaches 0 regardless of theta (because cos theta sin theta is bounded). The limit is 0.
Substitute: r^2 cos theta sin theta / r^2 = cos theta sin theta.
The r^2 terms cancel completely, leaving cos theta sin theta, which depends on the angle. At theta = 0 the value is 0; at theta = pi/4 the value is 1/2. The limit does not exist.
Substitute: r^3 cos theta sin^2 theta / (r^2 cos^2 theta + r^4 sin^4 theta). Factor r^2 from the denominator: r cos theta sin^2 theta / (cos^2 theta + r^2 sin^4 theta). As r approaches 0, this becomes 0 / cos^2 theta. The result depends on theta (it is 0 when cos theta is not zero, but undefined when theta = pi/2). The limit does not exist.
This is the same function that defeats all straight-line path tests. Polar coordinates catch it because the denominator does not simplify to a pure power of r.
For the limit of (x^2 + y^2) / (sqrt(x^2 + y^2 + 1) - 1), multiply numerator and denominator by the conjugate sqrt(x^2 + y^2 + 1) + 1.
The denominator becomes (x^2 + y^2 + 1) - 1 = x^2 + y^2. The (x^2 + y^2) factors cancel, leaving sqrt(x^2 + y^2 + 1) + 1. As (x, y) approaches (0, 0), this approaches sqrt(1) + 1 = 2. The limit is 2.
Polar coordinates only work for limits as (x, y) approaches (0, 0). If the target point is (a, b) with (a, b) not equal to (0, 0), you can shift the function: the limit of f(x, y) as (x, y) approaches (a, b) equals the limit of f(x + a, y + b) as (x, y) approaches (0, 0). Then polar coordinates apply to the shifted version.
In practice, most non-origin limits in this course can be evaluated by direct substitution or simple algebra without needing polar coordinates.
x = r\cos(\theta), \quad y = r\sin(\theta), \quad x^2 + y^2 = r^2The polar coordinate substitution. Approaching (0, 0) in Cartesian is equivalent to r approaching 0 from the right.
\sqrt{x^2 + y^2} \geq |x| \quad \Rightarrow \quad \frac{|x|}{\sqrt{x^2+y^2}} \leq 1The fundamental bounding inequality for Squeeze Theorem arguments in 2D.
\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}} = 0Proved by Squeeze Theorem or by polar coordinates (reduces to r cos theta sin theta, which approaches 0).
\lim_{(x,y)\to(0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1} = 2Evaluated by conjugate multiplication.
The Squeeze Theorem is used throughout applied mathematics and physics whenever you need to prove a quantity vanishes or converges without computing it exactly. In fluid dynamics, bounding arguments like these confirm that velocity fields are continuous at interior points. Polar coordinates are the natural language for any problem with radial symmetry: antenna radiation patterns, heat conduction from a point source, and gravitational or electrostatic potentials all reduce to functions of r and theta.
Students often think polar coordinates can prove a limit exists just because the expression simplifies. If theta survives in the final answer after r approaches 0, the limit does not exist. The simplification tells you something, but you must check whether the result still depends on angle.
A frequent mistake is forgetting that |x| / sqrt(x^2 + y^2) <= 1 only works because x^2 <= x^2 + y^2. Students sometimes try to use similar bounds without verifying the underlying inequality.
Some students apply polar coordinates to limits at points other than the origin without shifting first. Polar coordinates centre at the origin; if the target point is (1, 2), you must substitute x = 1 + r cos theta, y = 2 + r sin theta, or shift the function.
Students sometimes confuse "the limit equals 0 in polar form because r approaches 0" with "any expression multiplied by r goes to 0." This is only true if the rest of the expression (the part involving theta and possibly r) stays bounded. If the denominator also goes to 0, you need more care.
⚠️ Expect at least one "prove this limit exists" problem on the midterm. Path testing shows DNE; Squeeze Theorem or polar coordinates proves existence.
⚠️ The polar coordinate method is the fastest approach for most origin limits. If the problem says "evaluate," try polar first.
⚠️ When polar coordinates leave theta in the answer, state clearly that the limit depends on the direction of approach and therefore does not exist. Exams want that explicit statement.
⚠️ Conjugate multiplication comes up occasionally. Recognise the pattern: a denominator of the form sqrt(something) - constant.
⚠️ Know the shift trick for non-origin limits. It is not common on exams but appears as a harder problem.
True or false: The Squeeze Theorem can prove a limit does not exist. (False. It can only confirm existence by trapping the function between two bounds that agree.)
Fill in the blank: In polar coordinates, (x, y) approaching (0, 0) is equivalent to ____ approaching 0 from the right. (r)
True or false: If polar substitution gives a result that depends on theta, the limit exists and equals that expression. (False. If theta survives, the limit does not exist.)
Fill in the blank: The key bounding inequality for 2D Squeeze Theorem arguments is |x| / sqrt(x^2 + y^2) <= ____. (1)
True or false: Polar coordinates can be used directly for limits as (x, y) approaches (3, 5). (False. You must shift the function first so the limit is at the origin.)
Q: Use polar coordinates to evaluate the limit of xy / sqrt(x^2 + y^2) as (x, y) approaches (0, 0).
A: Substitute x = r cos theta, y = r sin theta. The expression becomes r^2 cos theta sin theta / r = r cos theta sin theta. As r approaches 0, this approaches 0 (since cos theta sin theta is bounded between -1 and 1). The limit is 0.
Q: Use polar coordinates to show that the limit of xy / (x^2 + y^2) as (x, y) approaches (0, 0) does not exist.
A: Substitute to get r^2 cos theta sin theta / r^2 = cos theta sin theta. The r terms cancel. The result depends on theta (e.g. it equals 0 at theta = 0 and 1/2 at theta = pi/4), so the limit does not exist.
Q: Evaluate the limit of (x^2 + y^2) / (sqrt(x^2 + y^2 + 1) - 1) as (x, y) approaches (0, 0).
A: Multiply numerator and denominator by the conjugate sqrt(x^2 + y^2 + 1) + 1. The denominator simplifies to x^2 + y^2, which cancels with the numerator, leaving sqrt(x^2 + y^2 + 1) + 1. As (x, y) approaches (0, 0), this approaches sqrt(1) + 1 = 2.
Q: Explain why polar coordinates alone cannot determine whether the limit of xy^2 / (x^2 + y^4) exists at the origin.
A: After polar substitution, the expression becomes r cos theta sin^2 theta / (cos^2 theta + r^2 sin^4 theta). As r approaches 0, this gives 0 / cos^2 theta. When cos theta is not zero, the value is 0, but when cos theta = 0 the expression is undefined (0/0). The polar form is inconclusive at theta = pi/2 and 3pi/2, which correspond to approach along the y-axis. This function requires the path x = y^2 to reveal the DNE.
This connects to path testing (the companion study notes), which is the first tool you reach for when evaluating multivariable limits. Path testing identifies limits that do not exist; the Squeeze Theorem and polar coordinates handle the ones that do.
Continuity of multivariable functions requires the limit to exist and to equal the function's value at that point. Every limit you prove with these techniques is a step toward establishing continuity.
Polar coordinates return later in the course for double integrals over circular regions, where the substitution x = r cos theta, y = r sin theta appears again with a Jacobian factor of r.
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