Limits and Continuity

Limits and Continuity Study Notes Institution: California State University, Northridge（Northridge, CA） Course: Math 250（Vector Calculus） Instructor: David Klein Instructor Time: Last Friday （I）Definition and Nature of Limits of Multivariable Functions (primarily binary functions, Professor David's lecture highlights) Strict Definition: Let f (x,y) be defined in the punctured neighborhood U°(P₀) of P₀(x₀,y₀) . If for any ε > 0, there exists δ > 0, such that when 0 < √[(x - x₀)² + (y - y₀)²] < δ , it always holds that f (x,y) - A < ε, then A is called the limit of f (x,y) as (x,y) → (x₀,y₀) , denoted as: lim [(x,y)→(x₀,y₀)] f (x,y) = A or lim [ρ→0] f (x₀ + ρcosθ, y₀ + ρsinθ) = A (where ρ = √[(x - x₀)² + (y - y₀)²]) The essence emphasized by the professor: Single-variable function limits are"approaching from both sides," while multivariable limits are"approaching from any path in the entire plane"—This is the core focus for determining non-existence of limits in course exams; it must be memorized: if there exists a single path where the limit differs, the overall limit does not exist. Notation standard: The classroom requires the use of lim [(x,y)→(x₀,y₀)] form first, which can be written as lim [ρ→0]after polar coordinate conversion (with the note ρ = √[(x - x₀)² + (y y₀)²]). (II) Techniques for solving and determining limits (high-frequency methods in the course) Three Major Solving Techniques (David Professor's Frequent Use in Classroom Examples) "Substitution Method": Only applicable when f (x,y) is continuous at (x₀,y₀) (to be emphasized later), directly substitute the coordinates for calculation. Example 1 (Basic Problem): Find lim [(x,y)→(2,3)] (3x² - 2xy + y²) Analysis: Since 3x² - 2xy + y² is a polynomial (a continuous function), direct substitution gives 3×2² - 2×2×3 + 3² = 12 - 12 + 9 = 9. Example Problem2 (including trigonometric functions): Find lim [(x,y)→(0,π/2)] (xsin y + ycos x) Analysis: sin y and cos x are both continuous functions, so substituting gives 0×sin (π/2) + (π/2)×cos0 = 0 + (π/2)×1 = π/2. Professor's Tip: The key to substitution method is“confirming the continuity of the function.” Elementary functions and their compositions satisfy continuity within their domains and can be used directly. 「Polar Coordinate Method」: A core method in the course, suitable for cases where the denominator contains x² + y² or when the numerator and denominator are homogeneous expressions. Operating steps: Let x = x₀ + ρcosθ, y = y₀ + ρsinθ (ρ → 0⁺), transform it into a single-variable limit about ρ. Key determination: If the result is independent of θ, then the limit exists; if it depends on θ, then the limit does not exist (must-know exam point). Example 1 (high-frequency exam question): Find lim [(x,y)→(0,0)] (x³ + y³)/(x² + y²) (original example). Solution: Cartesian coordinate conversion: x = ρcosθ，y = ρsinθ，x² + y² = ρ²，ρ → 0⁺。 Substitute into the original expression: (ρ³cos³θ + ρ³sin³θ)/ρ² = ρ(cos³θ + sin³θ). Scaling test: cos³θ + sin³θ ≤ 2（bounded），so ρ(...) ≤ 2ρ → limit is 0。 Example2（numerator degree higher than denominator）：Find lim [(x,y)→(0,0)] (x⁴ + 2x²y² + y⁴)/(x² + y²) Solution: Factor the numerator: x⁴ + 2x²y² + y⁴ = (x² + y²)²。 Conversion to polar coordinates: (ρ²)² / ρ² = ρ² → lim [ρ→0⁺] ρ² = 0 (independent of θ ). Example problem3 (determination of non-existent limit): Find lim [(x,y)→(0,0)] (x² - y²)/(x² + y²) Solution: Cartesian coordinate conversion: (ρ²cos²θ - ρ²sin²θ)/ρ² = cos2θ。 Result is related to θ (e.g. θ=0 gives1, θ=π/4 gives0), hence the limit does not exist. Professor's tip: In polar coordinate method, if the terms containing θ cannot be eliminated after conversion and the result is not unique, then the limit must not exist. This is a quick determination technique. "Squeeze Theorem": Common Inequalities xy ≤ (x² + y²)/2, sin t ≤ t, x ≤ √(x² + y²), need to be proficient in expansion and contraction techniques. Example1 (Basic Expansion and Contraction): Find lim [(x,y)→(0,0)] (x²y²)/(x² + y²) Solution: Using the inequality: x²y² ≤ (x² + y²)²/4（from xy ≤ (x² + y²)/2 square得）。 Scaling 0 ≤ (x²y²)/(x² + y²) ≤ (x² + y²)/4 → When (x,y)→(0,0) then, (x² + y²)/4 → 0。 By the squeeze theorem, the limit is 0。 Example 2 (including trigonometric functions): Find lim [(x,y)→(0,0)] (sin (xy))/(√(x² + y²)) Solution: Utilize sin (xy) ≤ xy and xy ≤ (x² + y²)/2。 By scaling 0 ≤ sin (xy)/√(x² + y²) ≤ (x² + y²)/(2√(x² + y²)) = √(x² + y²)/2 → approaching 0。 Thus, the limit is 0。 Professor's tip: The essence of the squeeze theorem is“finding two functions that approach the same value and sandwiching the target function between them,” common scaling inequalities should be memorized, and they can be directly applied in exams. Determination of Non-Existence of Limits (Common Exam Question Type) Method: Find two easily calculable paths (David Professor recommends prioritizing: ①coordinate axes y=0 or x=0；②straight line y=kx；③parabola y=kx²），if the limits are different, then the original limit does not exist. Example 1 (Classic Path Method Question): Determine whether the limit lim [(x,y)→(0,0)] (xy)/(x² + y²) exists (original example) Analysis: Along the x axis (y=0), the limit is 0, along y=x the limit is 1/2, so it does not exist. Example Problem2 (Parabolic Path): Determining lim [(x,y)→(0,0)] (x³)/(x⁴ + y²) exists Solution: Path1: Along x axis (y=0)→ lim [x→0] x³/x⁴ = lim [x→0] 1/x → infinity (limit does not exist), directly determine that the original limit does not exist. Additional Path2 (Alongy=x²): lim [x→0] x³/(x⁴ + x⁴) = lim [x→0] x³/(2x⁴) = lim [x→0] 1/(2x) → infinity, further verification. Example Problem3 (Lines with Different Slopes): Determination lim [(x,y)→(0,0)] (x²y)/(x⁴ + y²) exists or not Solution: Path1: Alongy=kx (k≠0) → lim [x→0] (x²・kx)/(x⁴ + k²x²) = lim [x→0] kx/(x² + k²) = 0. Path2: Along y=x² → lim [x→0] (x²・x²)/(x⁴ + x⁴) = lim [x→0] x⁴/(2x⁴) = 1/2. Two paths have different limits, so the original limit does not exist. Common Mistake (Emphasized Repeatedly by Professor): Cannot determine the existence of a limit solely based on one path or multiple parallel paths (such as y=2x, y=3x which are all straight lines), it must be ensured that"any path"converges to the same value; if the limit does not exist along a certain path (such as infinity or oscillation), then the original limit does not exist directly. (III) Definition and Determination Conditions of Continuity (Basic Exam Points) Three conditions (all are essential): f (x,y) is defined in some neighborhood of (x₀,y₀) ; lim [(x,y)→(x₀,y₀)] f (x,y) exists; limit value = function value, i.e., lim [(x,y)→(x₀,y₀)] f (x,y) = f (x₀,y₀). Common continuous functions: Composite functions of elementary functions (polynomials, trigonometric functions, exponential and logarithmic functions) summarized in class are continuous within their domains and can be directly used to find limits by substitution. Example Problems1 (Basic Continuity Determination): Discuss f (x,y) = { sin (xy)/(xy) , xy ≠ 0；1 , xy = 0 } at (0,0) (original example problem) Analysis: Satisfies the three conditions, hence continuous. Example Problems2 (Undefined Discontinuity): Discuss f (x,y) = (x² y²)/(x - y) at (1,1) continuity Solution: Domain: x ≠ y, hence f (x,y) is undefined at (1,1) (does not satisfy condition1)。 Conclusion: The function is discontinuous at (1,1) 。 Example 3 (Limit ≠ Discontinuity): Discuss the continuity of f (x,y) = { (x²y)/(x² + y²) , (x,y) ≠ (0,0)；2 , (x,y) = (0,0) } at (0,0) 。 Solution: Condition1: f(0,0)=2, well-defined; Condition2: By polar coordinates, we get lim [(x,y)→(0,0)] (x²y)/(x² + y²) = 0, the limit exists; Condition3: 0 ≠ 2, not satisfied. Conclusion: Discontinuous (removable discontinuity, if we let f (0,0)=0 then it becomes continuous). Example Problem4 (Continuity of Piecewise Functions): Discussion f (x,y) = { x² + y² , (x,y) ≠ (0,0)；0 , (x,y) = (0,0) } at (0,0) is continuous Solution: Condition1: f(0,0)=0, defined; Condition2: lim [(x,y)→(0,0)] (x² + y²) = 0, limit exists; Condition 3: 0 = 0, satisfies。 Conclusion: Continuous. Professor's Tip: The continuity of piecewise functions at boundary points is a key exam topic. It is essential to strictly verify the three conditions, paying particular attention to"limit exists"and"limit = function value"distinction. (IV) Relationship between limits and continuity (logical deduction focus) "Sufficiently unnecessary": Continuous → Limit exists (and equals the function value); but limit exists ≠ Continuous (may be undefined or limit ≠ function value). "Contrapositive (common in exams)": Limit does not exist → Discontinuous (can quickly eliminate continuity options). Example problem1 (Limit exists but not continuous): f (x,y) = { (x²y)/(x² + y²) , (x,y) ≠ (0,0)；1 , (x,y) = (0,0) } (original example), limit is0≠1, discontinuous. Example problem2 (Continuous if limit exists): f (x,y) = √(x² + y²) is continuous at (0,0) , so lim [(x,y)→(0,0)] √(x² + y²) = f (0,0) = 0. Example3 (limit does not exist, hence not continuous): f (x,y) = (xy)/(x² + y²) does not have a limit at (0,0) (original example), so the function is not continuous at (0,0) . Professor's Tip: Common in Exams"Finding Parameters Given Continuity" or "Determining Continuity," can use"the contrapositive" to quickly eliminate incorrect options —— if you can prove the limit does not exist, directly determine discontinuity. II. Key Examples from the Course (DavidProfessor's Teaching Style + Exam Question Types) Example 1: Finding Limits Using Polar Coordinates (High-frequency Exam Question Type) Problem: Find lim [(x,y)→(0,0)] (x³ + y³)/(x² + y²) (Original Example Problem) Solution: Cartesian coordinate conversion: x = ρcosθ，y = ρsinθ，x² + y² = ρ²，ρ → 0⁺。 Substitute into the original expression: (ρ³cos³θ + ρ³sin³θ)/ρ² = ρ(cos³θ + sin³θ). Scaling determination: cos³θ + sin³θ ≤ 2 (bounded), hence ρ(...) ≤ 2ρ → lim [ρ→0⁺] 2ρ = 0。 Conclusion: The original limit = 0。 Professor's comment: The core of polar coordinate method is "eliminating the homogeneous terms in the denominator", after transformation, use "bounded function × infinitesimal = infinitesimal" to solve, such types of problems account for a high proportion in exams. Example 2: Path method determining that the limit does not exist (must-examine exam question) Question: Determination lim [(x,y)→(0,0)] (xy)/(x² + y²) exists (original example) Solution: Path1: Along x axis (y=0) → lim [x→0] 0/x² = 0。 Path2: Along y=x → lim [x→0] x²/(2x²) = 1/2。 Conclusion: The limit does not exist. Common Mistake Tip: Avoid choosing complex paths; prioritize simple paths for quick determination. Example Problem3: Continuity Determination (Basic Must-Ask Question) Problem: Discuss f (x,y) = { sin (xy)/(xy) , xy ≠ 0；1 , xy = 0 } at (0,0) for continuity (original example problem) Solution: Condition1: f(0,0)=1, well-defined; Condition2: Lett=xy, t→0, lim [t→0] sin t/t = 1, the limit exists; Condition3: 1=1, satisfied. Conclusion: Continuous. Example Problem4: Comprehensive Application (Limits + Continuity to Find Parameters) Problem: Given f (x,y) = { (x² + y²)/(√(x² + y² + 1) - 1) , (x,y) ≠ (0,0)； k , (x,y) = (0,0) } is continuous at (0,0) find k value (original example problem) Solution: Continuous necessary condition: lim [(x,y)→(0,0)] f (x,y) = k。 Rationalization and simplification: Multiply both numerator and denominator by √(x² + y² + 1) + 1, resulting in √(x² + y² + 1) + 1。 Find the limit: → √1 + 1 = 2 → k=2。 Professor's Approach: Prioritize rationalizing expressions with square roots, then use continuity to find parameters, as this is a common comprehensive problem type in exams. Example 5: Comprehensive extension (limit + continuity + piecewise function) Problem: Let f (x,y) = { (x²sin y + y²cos x)/(x² + y²) , (x,y) ≠ (0,0)；a , (x,y) = (0,0) }，determine a