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HomeCalculus III: Multivariable CalculusStudy Guide: Limits and Continuity in Multivariable Calculus
Study Guide920 words

Study Guide: Limits and Continuity in Multivariable Calculus

Limits and Continuity

Limits and Continuity in Multivariable Calculus

This study guide covers the fundamental concepts of limits and continuity for functions of multiple variables, specifically focusing on functions of the form z=f(x,y)z = f(x, y)z=f(x,y). Unlike single-variable calculus, where a limit is approached from only two directions (left and right), multivariable limits must be consistent across an infinite number of paths.


Learning Objectives

After studying this material, you should be able to:

  • Define the limit of a function of two variables using the ϵ−δ\epsilon-\deltaϵ−δ formal definition.
  • Understand the concept of an open disk as the multivariable equivalent of an open interval.
  • Evaluate limits of multivariable functions using algebraic limit laws.
  • Prove that a limit does not exist using the Two-Path Test.
  • Verify the continuity of a function at a specific point (x0,y0)(x_0, y_0)(x0​,y0​).

The "Big Idea"

In single-variable calculus, the limit exists if the left-hand limit equals the right-hand limit. In multivariable calculus, the domain is a plane (or higher-dimensional space). For a limit lim⁡(x,y)→(x0,y0)f(x,y)=L\lim_{(x,y) \to (x_0,y_0)} f(x,y) = Llim(x,y)→(x0​,y0​)​f(x,y)=L to exist, the function must approach LLL regardless of which path in the xyxyxy-plane you take to reach (x0,y0)(x_0, y_0)(x0​,y0​). If you find even two paths that yield different results, the limit does not exist (DNE).


Key Terms & Glossary

  • Open Disk: A set of points (x,y)(x, y)(x,y) such that the distance to a center point (a,b)(a, b)(a,b) is less than a radius δ\deltaδ. Formally: (x−a)2+(y−b)2<δ\sqrt{(x-a)^2 + (y-b)^2} < \delta(x−a)2+(y−b)2​<δ.
    • Example: The region inside a circle, not including the boundary line.
  • ϵ−δ\epsilon-\deltaϵ−δ Definition: The formal proof framework for limits. For every error tolerance ϵ>0\epsilon > 0ϵ>0, there exists a distance δ>0\delta > 0δ>0 such that if the input is within δ\deltaδ of the target point, the output is within ϵ\epsilonϵ of the limit LLL.
  • Path-Dependence: A property where the limit value changes based on the curve (path) followed to reach the point.
    • Example: Approaching (0,0)(0,0)(0,0) along the x-axis vs. the line y=xy=xy=x.
  • Continuity: A function f(x,y)f(x, y)f(x,y) is continuous at (x0,y0)(x_0, y_0)(x0​,y0​) if the limit exists, the function is defined at that point, and the limit equals the function value.

Formula / Concept Box

LawFormula / Property
Limit Definition$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } 0 < \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta \implies
Sum Lawlim⁡(f+g)=lim⁡f+lim⁡g\lim (f+g) = \lim f + \lim glim(f+g)=limf+limg
Product Lawlim⁡(f⋅g)=(lim⁡f)⋅(lim⁡g)\lim (f \cdot g) = (\lim f) \cdot (\lim g)lim(f⋅g)=(limf)⋅(limg)
Quotient Lawlim⁡(f/g)=(lim⁡f)/(lim⁡g)\lim (f/g) = (\lim f) / (\lim g)lim(f/g)=(limf)/(limg), provided lim⁡g≠0\lim g \neq 0limg=0
Continuity Conditionlim⁡(x,y)→(x0,y0)f(x,y)=f(x0,y0)\lim_{(x,y) \to (x_0,y_0)} f(x,y) = f(x_0, y_0)lim(x,y)→(x0​,y0​)​f(x,y)=f(x0​,y0​)

Hierarchical Outline

  • I. Limits of Functions of Two Variables
    • A. From Intervals to Disks: In 1D we use ∣x−a∣<δ|x-a| < \delta∣x−a∣<δ; in 2D we use the distance formula for a disk.
    • B. The Formal Definition: Relates the proximity in the domain (disk) to the proximity in the range (z-axis).
  • II. Evaluating Limits
    • A. Direct Substitution: Applicable for polynomials and rational functions where the denominator is non-zero.
    • B. Limit Laws: Sum, Difference, Constant Multiple, Product, Quotient, Power, and Root laws.
  • III. Non-Existence of Limits
    • A. The Two-Path Test: If f(x,y)→L1f(x,y) \to L_1f(x,y)→L1​ along path C1C_1C1​ and f(x,y)→L2f(x,y) \to L_2f(x,y)→L2​ along path C2C_2C2​, and L1≠L2L_1 \neq L_2L1​=L2​, then the limit DNE.
    • B. Common Paths to Test: x=0x=0x=0 (y-axis), y=0y=0y=0 (x-axis), y=mxy=mxy=mx (lines), y=x2y=x^2y=x2 (parabolas).
  • IV. Continuity
    • A. Three Requirements: f(a,b)f(a,b)f(a,b) exists; lim⁡(x,y)→(a,b)f(x,y)\lim_{(x,y) \to (a,b)} f(x,y)lim(x,y)→(a,b)​f(x,y) exists; they are equal.
    • B. Composition of Continuous Functions: If ggg and hhh are continuous, f(x,y)=g(h(x,y))f(x,y) = g(h(x,y))f(x,y)=g(h(x,y)) is continuous.

Visual Anchors

Comparison: 1D vs 2D Approach

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Figure 1 — Mermaid diagram

Visualization of the Delta-Disk

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Figure 2 — TikZ diagram

Definition-Example Pairs

Definition: The Two-Path Test If a function f(x,y)f(x, y)f(x,y) approaches different values along two different paths as (x,y)(x, y)(x,y) approaches (x0,y0)(x_0, y_0)(x0​,y0​), then the limit of f(x,y)f(x, y)f(x,y) at that point does not exist.

[!TIP] Real-World Example: Think of a mountain pass. If you approach a peak from the North ridge, you might reach an altitude of 5,000ft. If you approach from the South valley, you might end at a cliff face at 4,800ft. Because the elevations don't meet at the same point from different directions, there is a "discontinuity" or no single "limit" height for that specific longitude/latitude coordinate.


Worked Examples

Example 1: Proving a Limit Does Not Exist

Problem: Show that lim⁡(x,y)→(0,0)x2−y2x2+y2\lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x^2 + y^2}lim(x,y)→(0,0)​x2+y2x2−y2​ does not exist.

Step 1: Test Path 1 (x-axis, y=0y=0y=0) lim⁡(x,0)→(0,0)x2−02x2+02=lim⁡x→0x2x2=1\lim_{(x,0) \to (0,0)} \frac{x^2 - 0^2}{x^2 + 0^2} = \lim_{x \to 0} \frac{x^2}{x^2} = 1lim(x,0)→(0,0)​x2+02x2−02​=limx→0​x2x2​=1

Step 2: Test Path 2 (y-axis, x=0x=0x=0) lim⁡(0,y)→(0,0)02−y202+y2=lim⁡y→0−y2y2=−1\lim_{(0,y) \to (0,0)} \frac{0^2 - y^2}{0^2 + y^2} = \lim_{y \to 0} \frac{-y^2}{y^2} = -1lim(0,y)→(0,0)​02+y202−y2​=limy→0​y2−y2​=−1

Conclusion: Since $$1 \neq -1$$, the limit depends on the path and therefore does not exist.

Example 2: Evaluating via Limit Laws

Problem: Evaluate lim⁡(x,y)→(1,2)(3x2y−x+5)\lim_{(x,y) \to (1,2)} (3x^2y - x + 5)lim(x,y)→(1,2)​(3x2y−x+5).

Solution: Because this is a polynomial, it is continuous everywhere. We can use direct substitution:

  1. 3(1)2(2)−(1)+53(1)^2(2) - (1) + 53(1)2(2)−(1)+5
  2. 3(2)−1+53(2) - 1 + 53(2)−1+5
  3. $6 - 1 + 5 = 10$ The limit is 10.

Checkpoint Questions

  1. What is the geometric difference between a neighborhood in 1D calculus and 2D calculus?
  2. If lim⁡(x,y)→(0,0)f(x,y)=5\lim_{(x,y) \to (0,0)} f(x,y) = 5lim(x,y)→(0,0)​f(x,y)=5 along the path y=xy=xy=x and along the path y=2xy=2xy=2x, does this prove the limit is 5? Why or why not?
  3. State the three conditions required for f(x,y)f(x,y)f(x,y) to be continuous at (a,b)(a,b)(a,b).
▶Click to expand answers
  1. In 1D, it is an open interval (a−δ,a+δ)(a-\delta, a+\delta)(a−δ,a+δ). In 2D, it is an open disk.
  2. No. Testing paths can only prove a limit does not exist. To prove it does exist, you must use limit laws, squeeze theorem, or the ϵ−δ\epsilon-\deltaϵ−δ definition, as there are infinitely many paths to check.
  3. (1) f(a,b)f(a,b)f(a,b) is defined; (2) The limit as (x,y)→(a,b)(x,y) \to (a,b)(x,y)→(a,b) exists; (3) The limit equals f(a,b)f(a,b)f(a,b).
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Loading Diagram...
Flowchart, left to right. Left Limit connects to a. Right Limit connects to Target. Path: y=0 connects to a,b. Path: x=0 connects to Target2. Path: y=mx connects to Target2. Path: y=x^2 connects to Target2.