📐 Concept diagram

06-02 - Limits and Continuity in ℝⁿ

Phase: 6 | Subject: 06-02 Prerequisites: 06-01-functions-of-several-variables.md (domain/range, level curves), 04-01-limits.md (single-variable limits), 04-02-continuity.md (single-variable continuity) Next subject: 06-03-partial-derivatives.md

Learning Objectives

By the end of this subject, you will be able to:

1. Define limits of multivariable functions using the ε-δ definition
2. Compute limits of f(x,y) as (x,y) → (a,b) using direct substitution and algebraic techniques
3. Apply the two-path test to prove a limit does NOT exist
4. Define continuity for functions of several variables
5. Identify common pitfalls: confusing path-dependence with existence

Core Content

The Limit Concept in ℝⁿ

⚠️ CRITICAL FOUNDATION: Multivariable limits are fundamentally different from single-variable limits. In 1D, you only approach from left and right. In ℝⁿ, you can approach along infinitely many paths — lines, parabolas, spirals — and the limit only exists if ALL paths give the same value.

For single-variable functions: "lim_{x→a} f(x) = L" means f(x) approaches L as x approaches a from either side (left and right limits must agree).

For f(x,y): (x,y) can approach (a,b) from infinitely many directions along infinitely many paths — not just left and right. The limit exists ONLY if f(x,y) approaches the SAME value L no matter which path (x,y) takes to (a,b).

Formal definition:

$lim_{(x,y)→(a,b)} f(x,y) = L
$

means: for every ε > 0, there exists δ > 0 such that

$0 < √((x-a)² + (y-b)²) < δ  ⇒  |f(x,y) - L| < ε
$

In words: f(x,y) can be made arbitrarily close to L by taking (x,y) sufficiently close to (a,b) — regardless of direction.

The distance in ℝ² is the Euclidean distance: d = √((x-a)² + (y-b)²).

Computing Limits That Exist

Direct Substitution

If f(x,y) is a "nice" combination of polynomials, exponentials, trig, etc. (i.e., continuous) at (a,b), just plug in:

Example: lim_{(x,y)→(1,2)} (3x²y + 2) = 3(1)²(2) + 2 = 6 + 2 = 8.

Example: lim_{(x,y)→(0,π)} eˣ cos(y) = e⁰ cos(π) = 1 · (-1) = -1.

Rational Functions — Check Denominator

Example: lim_{(x,y)→(1,2)} (x² + y²)/(x + y)

Denominator at (1,2): 1 + 2 = 3 ≠ 0. Direct substitution works: (1 + 4)/3 = 5/3.

Example: lim_{(x,y)→(0,0)} (x² - y²)/(x² + y²)

At (0,0), denominator = 0 → 0/0 indeterminate form. Cannot substitute directly. Must test paths (see below — this limit does NOT exist).

Algebraic Manipulation

Example: lim_{(x,y)→(0,0)} (x²y)/(x² + y²)

Strategy: Bound the function. Note that |x| ≤ √(x²+y²) and |y| ≤ √(x²+y²).

|f(x,y)| = |x²y|/(x² + y²). Since |x| ≤ √(x²+y²) and |y| ≤ √(x²+y²), we can bound:

Better approach: Use polar coordinates. Let x = r cos θ, y = r sin θ, where r = √(x²+y²) → 0.

f = (r² cos²θ)(r sin θ)/r² = r cos²θ sin θ.

As r → 0, f → 0 regardless of θ. Therefore the limit exists and equals 0.

Polar Coordinates Method

For limits at (0,0) with rational functions, substitute x = r cos θ, y = r sin θ:

Key observation: If the resulting expression factors as rᵏ·g(θ) where g(θ) is bounded, then the limit is 0 as r → 0.

Example: lim_{(x,y)→(0,0)} (x³)/(x² + y²) x = r cos θ, y = r sin θ: f = r³ cos³θ / r² = r cos³θ → 0 as r → 0. Limit = 0.

Example: lim_{(x,y)→(0,0)} (xy²)/(x² + y²) f = (r cos θ)(r² sin²θ)/r² = r cos θ sin²θ → 0 as r → 0. Limit = 0.

The Two-Path Test: Proving a Limit Does NOT Exist

If two different paths to (a,b) give different limits, the overall limit does not exist.

Choosing Test Paths

Common approach paths: 1. Along the x-axis: set y = 0, then let x → a. 2. Along the y-axis: set x = 0, then let y → b. 3. Along lines: y = mx (through origin) or y = m(x-a) + b. 4. Along parabolas: y = kx², x = ky². 5. Along other curves: y = x^(3/2), etc.

Classic Two-Path Examples

Example 1: lim_{(x,y)→(0,0)} (x² - y²)/(x² + y²)

Path 1: y = 0 → f(x,0) = x²/x² = 1. Limit along x-axis = 1.

Path 2: x = 0 → f(0,y) = -y²/y² = -1. Limit along y-axis = -1.

Since 1 ≠ -1, the limit does NOT exist.

Example 2: lim_{(x,y)→(0,0)} (xy)/(x² + y²)

Path 1: y = 0 → f(x,0) = 0/x² = 0. Path 2: x = 0 → f(0,y) = 0/y² = 0. Path 3 (critical): y = x → f(x,x) = x²/(2x²) = 1/2. Path 4: y = -x → f(x,-x) = -x²/(2x²) = -1/2.

Different paths give 0, 1/2, and -1/2 → limit does NOT exist.

Key insight: Just checking axes is not enough! Two paths agree but a third disagrees = no limit.

Example 3: lim_{(x,y)→(0,0)} (xy²)/(x² + y⁴)

Path 1: x = 0 → 0. Path 2: y = 0 → 0. Path 3: y = √x → f(x,√x) = (x·x)/(x² + x²) = x²/(2x²) = 1/2.

Limit does NOT exist.

More Sophisticated Paths

Example 4: lim_{(x,y)→(0,0)} (x²y)/(x⁴ + y²)

Along any line y = mx: f(x,mx) = mx³/(x⁴ + m²x²) = mx/(x² + m²) → 0 as x → 0.

All lines give limit 0. But try y = x²: f(x,x²) = x⁴/(x⁴ + x⁴) = 1/2.

So limit does NOT exist, even though ALL straight lines give 0! This shows that checking lines alone is insufficient.

Strategy Summary

Continuity in Multiple Variables

Definition: f(x,y) is continuous at (a,b) if: 1. f(a,b) is defined 2. lim_{(x,y)→(a,b)} f(x,y) exists 3. The limit equals f(a,b)

Equivalently: small changes in x and y produce small changes in f.

Continuity Theorems

• Polynomials in x, y are continuous everywhere on ℝ².
• Rational functions are continuous where denominator ≠ 0.
• Exponential, trigonometric, logarithmic functions are continuous on their domains.
• Compositions of continuous functions are continuous (on appropriate domains).
• Sums, products, quotients of continuous functions are continuous (quotients where denominator ≠ 0).

Example: f(x,y) = sin(x² + y²) is continuous everywhere (composition of sin with polynomial).

Example: f(x,y) = e^(xy)/(x² + y²) is continuous everywhere EXCEPT at (0,0) (denominator zero there).

Points of Discontinuity

Removable discontinuity: The limit exists but doesn't equal f(a,b) (or f(a,b) is undefined). Fix by redefining.

Essential discontinuity: The limit does not exist, no matter how you define f(a,b).

Example: f(x,y) = { (x² - y²)/(x² + y²) if (x,y) ≠ (0,0); 0 if (x,y) = (0,0) } This has an essential discontinuity at (0,0) — the limit doesn't exist.

Key Terms

• Compositions of continuous functions
• Polynomials
• Rational functions

Worked Examples

Example 1: Direct Evaluation

Problem: Compute lim_{(x,y)→(2,3)} (x²y + xy²).

Solution: Both x²y and xy² are polynomials, continuous everywhere. Direct substitution: (2)²(3) + (2)(3)² = 4·3 + 2·9 = 12 + 18 = 30.

Example 2: Two-Path Test — Limit Does Not Exist

Problem: Determine if lim_{(x,y)→(0,0)} (2xy)/(x² + y²) exists.

Solution:

Step 1: Check y = 0: f(x,0) = 0/x² = 0. Along x-axis, limit = 0.

Step 2: Check x = 0: f(0,y) = 0/y² = 0. Along y-axis, limit = 0. So far consistent.

Step 3: Check y = x: f(x,x) = 2x²/(2x²) = 1. Along y = x, limit = 1.

Step 4: Since 0 ≠ 1, the limit does NOT exist.

Example 3: Polar Coordinates — Limit Exists

Problem: Compute lim_{(x,y)→(0,0)} (x²y + xy²)/(x² + y²).

Solution:

Step 1: Convert to polar: x = r cos θ, y = r sin θ.

Step 2: Numerator: r³ cos²θ sin θ + r³ cos θ sin²θ = r³ cos θ sin θ (cos θ + sin θ).

Step 3: Denominator: r².

Step 4: f = r cos θ sin θ (cos θ + sin θ).

Step 5: As r → 0, f → 0 regardless of θ (since |cos θ sin θ(cos θ + sin θ)| ≤ 2 for all θ, bounded).

Step 6: The limit exists and equals 0.

Example 4: Sophisticated Paths

Problem: Show that lim_{(x,y)→(0,0)} (x³y)/(x⁶ + y²) does not exist.

Solution:

Step 1: Try lines y = mx: f(x,mx) = mx⁴/(x⁶ + m²x²) → divide numerator and denominator by x²: = mx²/(x⁴ + m²) → 0 as x → 0.

All lines give 0. But we must check further.

Step 2: Try y = x³: f(x,x³) = x³·x³/(x⁶ + x⁶) = x⁶/(2x⁶) = 1/2.

Step 3: Different paths give different limits (0 vs 1/2), so the limit does NOT exist.

Quiz

Q1: What does the concept of Polynomials primarily refer to in this subject?

A) A visual representation of Polynomials B) A historical anecdote about Polynomials C) A computational error related to Polynomials D) The definition and application of Polynomials

Correct: D)

• If you chose A: This is incorrect. Polynomials is defined as: the definition and application of polynomials. The other options describe different aspects that are not the primary focus.
• If you chose B: This is incorrect. Polynomials is defined as: the definition and application of polynomials. The other options describe different aspects that are not the primary focus.
• If you chose C: This is incorrect. Polynomials is defined as: the definition and application of polynomials. The other options describe different aspects that are not the primary focus.
• If you chose D: Polynomials is defined as: the definition and application of polynomials. The other options describe different aspects that are not the primary focus. Correct!

Q2: What is the primary purpose of Rational functions?

A) It is used only in advanced research contexts B) It is primarily a historical notation system C) It is used to rational functions in mathematical analysis D) It replaces all other methods in this domain

Correct: C)

• If you chose A: This is incorrect. Rational functions serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose B: This is incorrect. Rational functions serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose C: Rational functions serves the purpose described in the correct answer. The other options misrepresent its role. Correct!
• If you chose D: This is incorrect. Rational functions serves the purpose described in the correct answer. The other options misrepresent its role.

Q3: Which statement about Compositions of continuous functions is TRUE?

A) Compositions of continuous functions is mentioned only as a historical footnote B) Compositions of continuous functions is a fundamental concept covered in this subject C) Compositions of continuous functions is not related to this subject D) Compositions of continuous functions is an advanced topic beyond this subject's scope

Correct: B)

• If you chose A: This is incorrect. Compositions of continuous functions is a fundamental concept covered in this subject. This subject covers Compositions of continuous functions as part of its core content.
• If you chose B: Compositions of continuous functions is a fundamental concept covered in this subject. This subject covers Compositions of continuous functions as part of its core content. Correct!
• If you chose C: This is incorrect. Compositions of continuous functions is a fundamental concept covered in this subject. This subject covers Compositions of continuous functions as part of its core content.
• If you chose D: This is incorrect. Compositions of continuous functions is a fundamental concept covered in this subject. This subject covers Compositions of continuous functions as part of its core content.

Q4: Based on the worked examples in this subject, what is the correct result?

A) The inverse of the correct answer B) An unrelated numerical value C) 7. D) A different result from a common mistake

Correct: C)

• If you chose A: This is incorrect. The worked examples show that the result is 7.. The other options represent common errors.
• If you chose B: This is incorrect. The worked examples show that the result is 7.. The other options represent common errors.
• If you chose C: The worked examples show that the result is 7.. The other options represent common errors. Correct!
• If you chose D: This is incorrect. The worked examples show that the result is 7.. The other options represent common errors.

Q5: How are Compositions of continuous functions and The Limit Concept In ℝⁿ related?

A) Compositions of continuous functions is a special case of The Limit Concept In ℝⁿ B) Compositions of continuous functions is the inverse of The Limit Concept In ℝⁿ C) Compositions of continuous functions and The Limit Concept In ℝⁿ are completely unrelated topics D) Compositions of continuous functions and The Limit Concept In ℝⁿ are closely related concepts

Correct: D)

• If you chose A: This is incorrect. Both Compositions of continuous functions and The Limit Concept In ℝⁿ are covered in this subject as interconnected topics.
• If you chose B: This is incorrect. Both Compositions of continuous functions and The Limit Concept In ℝⁿ are covered in this subject as interconnected topics.
• If you chose C: This is incorrect. Both Compositions of continuous functions and The Limit Concept In ℝⁿ are covered in this subject as interconnected topics.
• If you chose D: Both Compositions of continuous functions and The Limit Concept In ℝⁿ are covered in this subject as interconnected topics. Correct!

Q6: What is a common pitfall when working with Computing Limits That Exist?

A) The main error with Computing Limits That Exist is using it when it is not needed B) A common mistake is confusing Computing Limits That Exist with a similar concept C) Computing Limits That Exist is always computed the same way in all contexts D) Computing Limits That Exist has no common misconceptions

Correct: B)

• If you chose A: This is incorrect. Students often confuse Computing Limits That Exist with similar-sounding or related concepts. Pay attention to the precise definitions.
• If you chose B: Students often confuse Computing Limits That Exist with similar-sounding or related concepts. Pay attention to the precise definitions. Correct!
• If you chose C: This is incorrect. Students often confuse Computing Limits That Exist with similar-sounding or related concepts. Pay attention to the precise definitions.
• If you chose D: This is incorrect. Students often confuse Computing Limits That Exist with similar-sounding or related concepts. Pay attention to the precise definitions.

Q7: When should you apply Direct Substitution?

A) Avoid Direct Substitution unless explicitly instructed B) Use Direct Substitution only in pure mathematics contexts C) Apply Direct Substitution to solve problems in this subject's domain D) Direct Substitution is not practically useful

Correct: C)

• If you chose A: This is incorrect. Direct Substitution is a practical tool used throughout this subject to solve relevant problems.
• If you chose B: This is incorrect. Direct Substitution is a practical tool used throughout this subject to solve relevant problems.
• If you chose C: Direct Substitution is a practical tool used throughout this subject to solve relevant problems. Correct!
• If you chose D: This is incorrect. Direct Substitution is a practical tool used throughout this subject to solve relevant problems.

Practice Problems

1. Compute lim_{(x,y)→(3,-1)} (x² + xy + y²).

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

3. Use polar coordinates to evaluate lim_{(x,y)→(0,0)} (x³ + y³)/(x² + y²).

4. Is f(x,y) = (x² + y²)/sin(x² + y²) continuous at (0,0)? Explain.

5. Show that lim_{(x,y)→(0,0)} (x⁴)/(x⁴ + y²) does not exist.

6. Determine where f(x,y) = ln(xy - 1) is continuous.

Summary

Key takeaways:

• Limits in ℝⁿ require the SAME limit along ALL possible paths — infinitely many
• Two-path test: find two paths with different limits to prove non-existence
• Always check beyond the axes; lines may all agree but a parabolic path may differ
• Polar coordinates (x = r cos θ, y = r sin θ) are powerful for proving limits exist at the origin
• Continuity in ℝⁿ: limit equals function value; polynomials and compositions of continuous functions are continuous on their domains

Pitfalls

• Assuming two-path agreement proves existence. Showing that limits along the x-axis and y-axis agree does NOT prove the limit exists. There are infinitely many other paths (lines y = mx, parabolas y = kx², cubics, spirals, etc.) and ALL must give the same limit.
• Stopping after checking only straight-line paths. For limits like lim_{(x,y)→(0,0)} (x²y)/(x⁴ + y²), every straight-line path through the origin gives limit 0, but the path y = x² gives 1/2. Checking only lines is insufficient — curved paths must also be considered.
• Using polar coordinates but ignoring θ-dependence. Substituting x = r cos θ, y = r sin θ and saying "as r → 0, f → 0" only works if the expression factors as rᵏ·g(θ) with g(θ) bounded and independent of r. If the limit depends on θ after taking r → 0, the limit does not exist.
• Forgetting to verify all three continuity conditions. A function is continuous at (a,b) only if: (1) f(a,b) is defined, (2) the limit exists, AND (3) the limit equals f(a,b). Checking only one or two conditions is a common oversight.
• Confusing continuity with differentiability or existence of partial derivatives. In ℝⁿ, a function can have partial derivatives at a point but not be continuous there — unlike the single-variable case where differentiability implies continuity. This counterintuitive fact catches many students off guard.

Next Steps

Next up: 06-03-partial-derivatives.md