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Limits and Continuity

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Uploaded on
November 16, 2025
Number of pages
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Written in
2025/2026
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Class notes
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David klein
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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:
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