Questions: Limits and Continuity in Multivariable Functions

Question 1

A student checks the limit of f(x,y) as (x,y)→(0,0) along five different paths and finds the value L = 0 in every case. Can they conclude the limit equals 0?

• A. Yes — checking five paths with consistent results is sufficient to establish the limit
• B. No — infinitely many paths remain unchecked; only the epsilon-delta definition or a bounding argument can prove the limit exists
• C. Yes — if the limit is the same along straight lines at all angles, it is the same along all paths
• D. Yes, provided the paths include both straight lines and at least one curved path

Answer: (B)

The path test can only disprove a limit. Two paths giving different values proves no limit exists, but checking finitely many consistent paths can never prove the limit exists, because infinitely many paths remain unchecked. A function can give 0 along every straight line through the origin and still fail to have a limit (e.g., returning a different value along y = x²). Proof of existence requires the epsilon-delta framework or a bounding argument that covers all paths at once.

Question 2

For f(x,y) = xy/(x² + y²), a student evaluates the limit at the origin along y = 0 and gets 0, then along y = x and gets 1/2. What is the correct conclusion?

• A. The limit is somewhere between 0 and 1/2; more paths are needed to pinpoint it
• B. The limit does not exist at the origin, because two paths give different limiting values
• C. The limit is 0, since y = 0 is the simpler and more natural path
• D. The limit is 1/2, since the path y = x is more general than y = 0

Answer: (B)

If two paths to the same point give different limiting values, the limit does not exist — full stop. A limit requires the function to approach the same value L no matter how you approach the point. Different results along y = 0 and y = x prove that no single value L satisfies the definition for all paths. There is no 'most correct' path among conflicting ones.

Question 3

In multivariable calculus, if two different paths to a point (a, b) yield different limiting values for f(x, y), then the limit of f at (a, b) does not exist.

• T. True
• F. False

Answer: True

The limit lim_{(x,y)→(a,b)} f(x,y) = L must hold for every possible path of approach simultaneously. If even one path gives a different value, the limit does not exist. This is the path test: a sufficient condition for non-existence. It cannot prove existence, because you cannot check all paths by testing finitely many.

Question 4

Showing that lim_{(x,y)→(0,0)} f(x,y) = 0 along nearly every straight line through the origin is sufficient to prove the limit equals 0.

• T. True
• F. False

Answer: False

Straight lines form only a subset of the infinitely many paths to the origin. A function can give the limit 0 along every line y = mx yet approach a different value along a curved path such as y = x². The limit exists only if the function approaches the same value along every conceivable path — including parabolas, spirals, oscillating curves, and others.

Question 5

Explain why the path test in multivariable limits can disprove the existence of a limit but cannot prove that a limit exists.

Model answer: The path test checks finitely many paths, but infinitely many paths approach any point. Disproving existence requires only one counterexample — a single path giving a different value. Proving existence requires showing the function approaches L along every path, which cannot be completed by checking examples. Proof requires the epsilon-delta definition or a bounding argument (such as polar coordinates) that covers all paths simultaneously.

This asymmetry is fundamental: existence claims require universal quantification (for all paths), so a proof must handle all cases at once. Non-existence claims require only one counterexample. The polar coordinate approach — setting x = r cosθ, y = r sinθ and showing the expression approaches L as r → 0 independently of θ — is the standard tool for proving existence, because r → 0 captures all paths simultaneously.