Calculus Volume 3 · Chapter 4 · Section 4.2
Limits and Continuity
On a line you reach a point from two sides. In the plane you can arrive from every direction — and a limit has to agree on all of them.
Colour key
key idea
· drag a 3D figure to turn it · page links open the textbook
Level 1 · see it What a limit in the plane asks before any law
Does the height settle at the origin?
f ( x , y ) = 2 x y 3 x 2 + y 2 f(x,y)=\dfrac{2xy}{3x^2+y^2} f ( x , y ) = 3 x 2 + y 2 2 x y ( 0 , 0 ) (0,0) ( 0 , 0 ) Example 4.9 ). Walk toward the origin two ways:
Along y = 0 y=0 y = 0 f = 0 f=0 f = 0 at every step.
Along y = x y=x y = x f = 1 2 f=\tfrac12 f = 2 1 at every step.
Two ways in, two heights: there is no single number to approach.
walk toward (0, 0) 20%
walk in
The graph is a fan of level lines through the z z z
Close means inside a small disc
δ disc · p317
( x − a ) 2 + ( y − b ) 2 < δ 2 (x-a)^2+(y-b)^2<\delta^2 ( x − a ) 2 + ( y − b ) 2 < δ 2 The plane's version of a − δ < x < a + δ a-\delta<x<a+\delta a − δ < x < a + δ
Limit · p317
lim ( x , y ) → ( a , b ) f ( x , y ) = L \lim_{(x,y)\to(a,b)}f(x,y)=L ( x , y ) → ( a , b ) lim f ( x , y ) = L For every ε > 0 \varepsilon>0 ε > 0 δ > 0 \delta>0 δ > 0 ( a , b ) (a,b) ( a , b ) ∣ f ( x , y ) − L ∣ < ε |f(x,y)-L|<\varepsilon ∣ f ( x , y ) − L ∣ < ε
Every point of the disc — not just the points on one line.
Pick an ε band around L L L Figure 4.15 ).
Infinitely many ways in
On a line, x → a x\to a x → a ( x , y ) → ( a , b ) (x,y)\to(a,b) ( x , y ) → ( a , b )
Example 4.9b (p321 ): f = 4 x y 2 x 2 + 3 y 4 f=\dfrac{4xy^2}{x^2+3y^4} f = x 2 + 3 y 4 4 x y 2 every line y = k x y=kx y = k x parabola x = y 2 x=y^2 x = y 2 p322 ).
Paths that disagree prove there is no limit. Paths that agree — even all the lines — prove nothing.
The level curves are parabolas x = k y 2 x=ky^2 x = k y 2 4 k / ( k 2 + 3 ) 4k/(k^2+3) 4 k / ( k 2 + 3 )
Continuous: no hole, no split
f f f ( a , b ) (a,b) ( a , b ) p325 )
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,
and the two are equal.
A missing value can be filled in. A limit that splits by direction cannot be repaired.
define f(0, 0) = 1
sin ( x 2 + y 2 ) / ( x 2 + y 2 ) → 1 \sin(x^2+y^2)/(x^2+y^2)\to1 sin ( x 2 + y 2 ) / ( x 2 + y 2 ) → 1
One picture
Level curves of 2 x y / ( 3 x 2 + y 2 ) 2xy/(3x^2+y^2) 2 x y / ( 3 x 2 + y 2 ) − 0.58 -0.58 − 0.58 0.58 0.58 0.58 ( 0 , 0 ) (0,0) ( 0 , 0 )
A limit exists only when every way in arrives at the same height.
Level 2 · compute it Laws, boundaries, continuity and four labs
Limit laws: substitute when nothing breaks
Theorem 4.1 (p318 ), with lim f = L \lim f=L lim f = L lim g = M \lim g=M lim g = M ( a , b ) (a,b) ( a , b )
law limit constant, identity lim c = c , lim x = a , lim y = b \lim c=c,\quad \lim x=a,\quad \lim y=b lim c = c , lim x = a , lim y = b sum, difference, multiple lim ( f ± g ) = L ± M , lim c f = c L \lim(f\pm g)=L\pm M,\quad \lim cf=cL lim ( f ± g ) = L ± M , lim c f = c L product, quotient lim f g = L M , lim f / g = L / M ( M ≠ 0 ) \lim fg=LM,\quad \lim f/g=L/M\ \ (M\ne0) lim f g = L M , lim f / g = L / M ( M = 0 ) power, root lim f n = L n , lim f n = L n \lim f^n=L^n,\quad \lim\sqrt[n]{f}=\sqrt[n]{L} lim f n = L n , lim n f = n L
Polynomials, and quotients whose bottom is not 0, take their limit by substitution. A 0/0 is where the work starts.
Examples 4.8–4.13
Example given result 4.8a · p319 x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 x^2-2xy+3y^2-4x+3y-6 x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 ( 2 , − 1 ) (2,-1) ( 2 , − 1 ) − 6 -6 − 6 4.8b · p320 2 x + 3 y 4 x − 3 y \dfrac{2x+3y}{4x-3y} 4 x − 3 y 2 x + 3 y ( 2 , − 1 ) (2,-1) ( 2 , − 1 ) bottom → 11 ≠ 0 \to11\ne0 → 11 = 0 1 11 \tfrac1{11} 11 1 4.9a · p320 2 x y 3 x 2 + y 2 \dfrac{2xy}{3x^2+y^2} 3 x 2 + y 2 2 x y ( 0 , 0 ) (0,0) ( 0 , 0 ) y = 0 y=0 y = 0 y = x y=x y = x 1 2 \tfrac12 2 1 4.9b · p321 4 x y 2 x 2 + 3 y 4 \dfrac{4xy^2}{x^2+3y^4} x 2 + 3 y 4 4 x y 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) every line: 0; x = y 2 x=y^2 x = y 2 4.10 · p323 25 − x 2 − y 2 \sqrt{25-x^2-y^2} 25 − x 2 − y 2 ( 4 , 3 ) (4,3) ( 4 , 3 ) 0 0 0 4.11 · p325 3 x + 2 y x + y + 1 \dfrac{3x+2y}{x+y+1} x + y + 1 3 x + 2 y ( 5 , − 3 ) (5,-3) ( 5 , − 3 ) f ( 5 , − 3 ) = 3 = f(5,-3)=3= f ( 5 , − 3 ) = 3 = 4.12 · p326 4 x 3 y 2 4x^3y^2 4 x 3 y 2 cos ( 4 x 3 y 2 ) \cos(4x^3y^2) cos ( 4 x 3 y 2 ) continuous everywhere 4.13 · p327 x 2 y − 3 z 2 x + 5 y − z \dfrac{x^2y-3z}{2x+5y-z} 2 x + 5 y − z x 2 y − 3 z ( 4 , 1 , − 3 ) (4,1,-3) ( 4 , 1 , − 3 ) bottom 16, top 25: 25 16 \tfrac{25}{16} 16 25
Substitute first. A 0/0 means: try two paths to disprove, or simplify, bound or use polar coordinates to prove.
Interior, boundary, open, closed
Points · p322
Interior: some δ disc around it lies inside S S S Boundary: every δ disc holds points of S S S
Sets · p323
Open: every point is interior. Closed: contains its boundary. Region: open, connected, nonempty.
At a boundary point of the domain, only the points inside the domain count (Example 4.10 ): 25 − x 2 − y 2 → 0 \sqrt{25-x^2-y^2}\to0 25 − x 2 − y 2 → 0 ( 4 , 3 ) (4,3) ( 4 , 3 )
Figure 4.17 (p322 ): ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) ( 2 , 3 ) (2,3) ( 2 , 3 )
Lab 1 · approach along a path
Try: Example 4.9a, turn the line — the limit changes with θ. Example 4.9b: every line gives 0; switch to the parabola with k = 1 . Exercise 95: nothing you try disagrees.
Lab 3 · continuous or not?
at (0, 0)
Exercise 96
Exercise 95
Exercise 97
Exercise 103
condition here 1 · f ( 0 , 0 ) f(0,0) f ( 0 , 0 ) 2 · the limit exists 3 · they are equal
define f(0, 0) = 1
Three or more variables · Lab 4
δ ball · p327
( x − x 0 ) 2 + ( y − y 0 ) 2 + ( z − z 0 ) 2 < δ 2 (x-x_0)^2+(y-y_0)^2+(z-z_0)^2<\delta^2 ( x − x 0 ) 2 + ( y − y 0 ) 2 + ( z − z 0 ) 2 < δ 2 The same definition and laws; the disc becomes a ball.
Example 4.13 · p327
lim ( x , y , z ) → ( 4 , 1 , − 3 ) x 2 y − 3 z 2 x + 5 y − z = 25 16 \lim_{(x,y,z)\to(4,1,-3)}\frac{x^2y-3z}{2x+5y-z}=\frac{25}{16} ( x , y , z ) → ( 4 , 1 , − 3 ) lim 2 x + 5 y − z x 2 y − 3 z = 16 25 Bottom → 16 ≠ 0 \to16\ne0 → 16 = 0 → 25 \to25 → 25
δ · ball radius 1.5
grow the ball
The grey plane 2 x + 5 y − z = 0 2x+5y-z=0 2 x + 5 y − z = 0 16 / 30 ≈ 2.92 16/\sqrt{30}\approx2.92 16/ 30 ≈ 2.92
Common mistakes
Mistake Result Fix Two paths agree, so the limit exists Example 4.9b: every line gives 0, still no limit agreement proves nothing; prove with a bound Trying only lines y = k x y=kx y = k x missing x = y 2 x=y^2 x = y 2 try curves too, or polar coordinates "0/0, so no limit" sin ( x 2 + y 2 ) / ( x 2 + y 2 ) → 1 \sin(x^2+y^2)/(x^2+y^2)\to1 sin ( x 2 + y 2 ) / ( x 2 + y 2 ) → 1 0/0 means work to do Quotient law with a bottom tending to 0 a wrong or meaningless value check the bottom first: 11 in Example 4.8b Continuity from the limit alone x 2 y / ( x 2 + y 2 ) x^2y/(x^2+y^2) x 2 y / ( x 2 + y 2 ) all three conditions Polar limit that still depends on θ 2 cos θ sin θ / ( 3 cos 2 θ + sin 2 θ ) 2\cos\theta\sin\theta/(3\cos^2\theta+\sin^2\theta) 2 cos θ sin θ / ( 3 cos 2 θ + sin 2 θ ) a θ left over means no limit
Practice and answers
Exercises from p328 ; the book's key: p861 –862 , p866 –867 .
Checkpoints 4.6–4.11
# task answer 4.6 lim x 2 − y y 2 + x − 1 3 \lim\sqrt[3]{\dfrac{x^2-y}{y^2+x-1}} lim 3 y 2 + x − 1 x 2 − y ( 5 , − 2 ) (5,-2) ( 5 , − 2 ) 27 / 8 3 = 3 2 \sqrt[3]{27/8}=\tfrac32 3 27/8 = 2 3 4.7 ( x − 2 ) ( y − 1 ) ( x − 2 ) 2 + ( y − 1 ) 2 \dfrac{(x-2)(y-1)}{(x-2)^2+(y-1)^2} ( x − 2 ) 2 + ( y − 1 ) 2 ( x − 2 ) ( y − 1 ) ( 2 , 1 ) (2,1) ( 2 , 1 ) along y − 1 = k ( x − 2 ) y-1=k(x-2) y − 1 = k ( x − 2 ) k 1 + k 2 \tfrac{k}{1+k^2} 1 + k 2 k 4.8 29 − x 2 − y 2 \sqrt{29-x^2-y^2} 29 − x 2 − y 2 ( 5 , − 2 ) (5,-2) ( 5 , − 2 ) 0 0 0 4.9 26 − 2 x 2 − y 2 \sqrt{26-2x^2-y^2} 26 − 2 x 2 − y 2 ( 2 , − 3 ) (2,-3) ( 2 , − 3 ) f ( 2 , − 3 ) = 9 = 3 = f(2,-3)=\sqrt{9}=3= f ( 2 , − 3 ) = 9 = 3 = 4.10 2 x 2 y 3 + 3 2x^2y^3+3 2 x 2 y 3 + 3 ( 2 x 2 y 3 + 3 ) 4 (2x^2y^3+3)^4 ( 2 x 2 y 3 + 3 ) 4 products, sums, a composition: continuous everywhere 4.11 13 − x 2 − 2 y 2 + z 2 \sqrt{13-x^2-2y^2+z^2} 13 − x 2 − 2 y 2 + z 2 ( 4 , − 1 , 3 ) (4,-1,3) ( 4 , − 1 , 3 ) 4 = 2 \sqrt4=2 4 = 2
Exercises
# task answer 61 5 x 2 y x 2 + y 2 \dfrac{5x^2y}{x^2+y^2} x 2 + y 2 5 x 2 y ( 1 , 2 ) (1,2) ( 1 , 2 ) 2 2 2 63 4 x 2 + 10 y 2 + 4 4 x 2 − 10 y 2 + 6 \dfrac{4x^2+10y^2+4}{4x^2-10y^2+6} 4 x 2 − 10 y 2 + 6 4 x 2 + 10 y 2 + 4 ( 0 , 0 ) (0,0) ( 0 , 0 ) 2 3 \tfrac23 3 2 65 y 2 sin x x \dfrac{y^2\sin x}{x} x y 2 sin x ( 0 , 1 ) (0,1) ( 0 , 1 ) 1 1 1 67 y tan x y + 1 \dfrac{y\tan x}{y+1} y + 1 y tan x ( π / 4 , 1 ) (\pi/4,1) ( π /4 , 1 ) 1 2 \tfrac12 2 1 69 1 x − 5 y \tfrac1x-\tfrac5y x 1 − y 5 ( 2 , 5 ) (2,5) ( 2 , 5 ) − 1 2 -\tfrac12 − 2 1 71 e − x 2 − y 2 e^{-x^2-y^2} e − x 2 − y 2 ( 4 , 4 ) (4,4) ( 4 , 4 ) e − 32 e^{-32} e − 32 73 x 2 y 3 − x 3 y 2 + 3 x + 2 y x^2y^3-x^3y^2+3x+2y x 2 y 3 − x 3 y 2 + 3 x + 2 y ( 1 , 2 ) (1,2) ( 1 , 2 ) 11 11 11 75 x y + 1 x 2 + y 2 + 1 \dfrac{xy+1}{x^2+y^2+1} x 2 + y 2 + 1 x y + 1 ( 0 , 0 ) (0,0) ( 0 , 0 ) 1 1 1 77 ln ( x 2 + y 2 ) \ln(x^2+y^2) ln ( x 2 + y 2 ) ( 0 , 0 ) (0,0) ( 0 , 0 ) does not exist: → − ∞ \to-\infty → − ∞ 79 a boundary point of R R R one whose every open disc holds points inside and outside R R R 81 x 4 − 4 y 4 x 2 + 2 y 2 \dfrac{x^4-4y^4}{x^2+2y^2} x 2 + 2 y 2 x 4 − 4 y 4 ( 0 , 0 ) (0,0) ( 0 , 0 ) 0 0 0 x 2 − 2 y 2 x^2-2y^2 x 2 − 2 y 2 83 x 2 − x y x − y \dfrac{x^2-xy}{\sqrt x-\sqrt y} x − y x 2 − x y ( 0 , 0 ) (0,0) ( 0 , 0 ) 0 0 0 x ( x + y ) x(\sqrt x+\sqrt y) x ( x + y ) 85 x 2 − y 2 − z 2 x 2 + y 2 − z 2 \dfrac{x^2-y^2-z^2}{x^2+y^2-z^2} x 2 + y 2 − z 2 x 2 − y 2 − z 2 does not exist: x x x y y y − 1 -1 − 1 87 x y + y 3 x 2 + y 2 \dfrac{xy+y^3}{x^2+y^2} x 2 + y 2 x y + y 3 ( 0 , 0 ) (0,0) ( 0 , 0 ) does not exist: axes 0, y = 2 x y=2x y = 2 x 2 5 \tfrac25 5 2 89 x 2 y x 4 + y 2 \dfrac{x^2y}{x^4+y^2} x 4 + y 2 x 2 y ( 0 , 0 ) (0,0) ( 0 , 0 ) does not exist: axes 0, y = x 2 y=x^2 y = x 2 1 2 \tfrac12 2 1 91 ln ( x + y ) \ln(x+y) ln ( x + y ) continuous where y > − x y>-x y > − x 93 1 / ( x y ) 1/(xy) 1/ ( x y ) continuous where x ≠ 0 x\ne0 x = 0 y ≠ 0 y\ne0 y = 0 95 x 2 y x 2 + y 2 \dfrac{x^2y}{x^2+y^2} x 2 + y 2 x 2 y f ( 0 , 0 ) = 0 f(0,0)=0 f ( 0 , 0 ) = 0 continuous at ( 0 , 0 ) (0,0) ( 0 , 0 ) ∣ f ∣ ≤ ∣ y ∣ |f|\le|y| ∣ f ∣ ≤ ∣ y ∣ 97 x 2 − y 2 x 2 + y 2 \dfrac{x^2-y^2}{x^2+y^2} x 2 + y 2 x 2 − y 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) discontinuous: no limit, no value 99 arctan x y 2 x + y \arctan\dfrac{xy^2}{x+y} arctan x + y x y 2 continuous except on y = − x y=-x y = − x 101 x 2 + y 2 − 2 z 2 x^2+y^2-2z^2 x 2 + y 2 − 2 z 2 continuous at every point of space 103 1 x 2 + y 2 \dfrac1{x^2+y^2} x 2 + y 2 1 ( 0 , 0 ) (0,0) ( 0 , 0 ) no limit: grows without bound (Lab 3) 107 sin x 2 + y 2 x 2 + y 2 \dfrac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} x 2 + y 2 sin x 2 + y 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) 1 1 1 sin r / r \sin r/r sin r / r 109 f ( g ( x , y ) ) f(g(x,y)) f ( g ( x , y )) f ( t ) = 1 / t f(t)=1/t f ( t ) = 1/ t g = 2 x − 5 y g=2x-5y g = 2 x − 5 y continuous off the line 2 x − 5 y = 0 2x-5y=0 2 x − 5 y = 0 111 lim h → 0 f ( 1 + h , y ) − f ( 1 , y ) h \lim_{h\to0}\frac{f(1+h,y)-f(1,y)}{h} lim h → 0 h f ( 1 + h , y ) − f ( 1 , y ) f = x 2 − 4 y f=x^2-4y f = x 2 − 4 y 2 2 2
Level 3 · why it works Proving and disproving limits short arguments
Polar coordinates settle it
Put x = r cos θ x=r\cos\theta x = r cos θ y = r sin θ y=r\sin\theta y = r sin θ ( x , y ) → ( 0 , 0 ) (x,y)\to(0,0) ( x , y ) → ( 0 , 0 ) r → 0 r\to0 r → 0 θ \theta θ Exercise 107 ).
2 x y 3 x 2 + y 2 = 2 cos θ sin θ 3 cos 2 θ + sin 2 θ \frac{2xy}{3x^2+y^2}=\frac{2\cos\theta\sin\theta}{3\cos^2\theta+\sin^2\theta} 3 x 2 + y 2 2 x y = 3 cos 2 θ + sin 2 θ 2 cos θ sin θ No r r r
∣ x 2 y x 2 + y 2 ∣ = r cos 2 θ ∣ sin θ ∣ ≤ r → 0 \Bigl|\frac{x^2y}{x^2+y^2}\Bigr|=r\cos^2\theta\,|\sin\theta|\le r\to0 x 2 + y 2 x 2 y = r cos 2 θ ∣ sin θ ∣ ≤ r → 0 A bound in r r r
A limit is proved by a bound that shrinks with r r r θ \theta θ
Why every line is not enough
On the parabola x = k y 2 x=ky^2 x = k y 2
4 x y 2 x 2 + 3 y 4 ∣ x = k y 2 = 4 k k 2 + 3 , largest 2 3 ≈ 1.15 at k = 3 . \frac{4xy^2}{x^2+3y^4}\Big|_{x=ky^2}=\frac{4k}{k^2+3},\qquad \text{largest } \frac{2}{\sqrt3}\approx1.15 \text{ at } k=\sqrt3 . x 2 + 3 y 4 4 x y 2 x = k y 2 = k 2 + 3 4 k , largest 3 2 ≈ 1.15 at k = 3 . A line y = m x y=mx y = m x k = 1 / ( m 2 x ) k=1/(m^2x) k = 1/ ( m 2 x ) k → ∞ k\to\infty k → ∞ x → 0 x\to0 x → 0 4 k / ( k 2 + 3 ) → 0 4k/(k^2+3)\to0 4 k / ( k 2 + 3 ) → 0
Every line slips toward the flat parabolas at height 0; the ridge at height 1 is never on a line.
Why the sum law holds
Given ε \varepsilon ε δ 1 \delta_1 δ 1 ∣ f − L ∣ < ε / 2 |f-L|<\varepsilon/2 ∣ f − L ∣ < ε /2 δ 2 \delta_2 δ 2 ∣ g − M ∣ < ε / 2 |g-M|<\varepsilon/2 ∣ g − M ∣ < ε /2 p319 ). Inside the disc of radius δ = min ( δ 1 , δ 2 ) \delta=\min(\delta_1,\delta_2) δ = min ( δ 1 , δ 2 )
∣ f + g − ( L + M ) ∣ ≤ ∣ f − L ∣ + ∣ g − M ∣ < ε . |f+g-(L+M)|\le|f-L|+|g-M|<\varepsilon . ∣ f + g − ( L + M ) ∣ ≤ ∣ f − L ∣ + ∣ g − M ∣ < ε . The identity law: ∣ x − a ∣ ≤ ( x − a ) 2 + ( y − b ) 2 < δ |x-a|\le\sqrt{(x-a)^2+(y-b)^2}<\delta ∣ x − a ∣ ≤ ( x − a ) 2 + ( y − b ) 2 < δ δ = ε \delta=\varepsilon δ = ε
Each law is the one-variable proof with ∣ x − a ∣ |x-a| ∣ x − a ∣
Continuity survives composition
Theorems 4.2–4.4 (p326 ): if g g g ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) f f f g ( x 0 , y 0 ) g(x_0,y_0) g ( x 0 , y 0 ) f ∘ g f\circ g f ∘ g g g g f f f
Example 4.12: 4 x 3 4x^3 4 x 3 y 2 y^2 y 2 4 x 3 y 2 4x^3y^2 4 x 3 y 2 cos \cos cos cos ( 4 x 3 y 2 ) \cos(4x^3y^2) cos ( 4 x 3 y 2 )
Build a function from continuous pieces and it stays continuous — except where a piece is undefined.
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Calculus Volume 3 §4.2 Chapter 4 · Differentiation of Functions of Several Variables
Limits and Continuity
In the plane a point can be reached from every direction — a limit has to agree on all of them.
Find limits by the laws, disprove them with two paths, and check continuity as “limit = value”.
4.2.1 Calculate the limit of a function of two variables. 4.2.2 Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach. 4.2.3 State the conditions for continuity of a function of two variables. 4.2.4 Verify the continuity of a function of two variables at a point. 4.2.5 Calculate the limit of a function of three or more variables and verify the continuity of the function at a point.
Prereq: one-variable ε–δ limit; three continuity conditions Ask: "lim (x² − 4)/(x − 2) as x → 2?" → 4Ask: "Is |x|/x continuous at 0?" → no, −1 vs 1
Limit of a Function of Two Variables geometry Figure 4.14
In the plane, “close to ( a , b ) (a,b) ( a , b )
δ disk around ( a , b ) (a,b) ( a , b ) ( x , y ) (x,y) ( x , y ) ( x − a ) 2 + ( y − b ) 2 < δ 2 (x-a)^2+(y-b)^2<\delta^2 ( x − a ) 2 + ( y − b ) 2 < δ 2
One variable: a − δ < x < a + δ a-\delta<x<a+\delta a − δ < x < a + δ
Two variables: a whole disk around ( a , b ) (a,b) ( a , b )
… smaller δ, smaller disk — never a line.
Geometry open disk of radius δ
Algebra · p317 ( x − a ) 2 + ( y − b ) 2 < δ 2 (x-a)^2+(y-b)^2<\delta^2 ( x − a ) 2 + ( y − b ) 2 < δ 2
Ask: "Points within 0.5 of (2, 1)?" → a disk, not a squareFigure 4.14 is this disk
Limit of a Function of Two Variables bridge Definition · Figure 4.15
lim f = L \lim f=L lim f = L L L L For every ε > 0 \varepsilon>0 ε > 0 δ > 0 \delta>0 δ > 0 ∣ f ( x , y ) − L ∣ < ε |f(x,y)-L|<\varepsilon ∣ f ( x , y ) − L ∣ < ε 0 < ( x − a ) 2 + ( y − b ) 2 < δ 0<\sqrt{(x-a)^2+(y-b)^2}<\delta 0 < ( x − a ) 2 + ( y − b ) 2 < δ
You pick ε: the band L − ε L-\varepsilon L − ε L + ε L+\varepsilon L + ε
I answer with δ: the disk’s values stay in the band.
Smaller ε, smaller δ — always answerable ⇒ limit L L L
Geometry lifted δ disk inside the ε band
Algebra · p317 0 < dist < δ 0<\text{dist}<\delta 0 < dist < δ ⇒ ∣ f − L ∣ < ε \Rightarrow|f-L|<\varepsilon ⇒ ∣ f − L ∣ < ε
Say it as a game: you choose ε, I must answer δ ( a , b ) (a,b) ( a , b ) 0 < 0< 0 < Drag the figure to see the band from the side
Limit of a Function of Two Variables algebra Theorem 4.1
Limits pass through sums, products, quotients, powers and roots
If lim f = L \lim f=L lim f = L lim g = M \lim g=M lim g = M L L L M M M M ≠ 0 M\ne0 M = 0 L ≥ 0 L\ge0 L ≥ 0
lim c = c , lim x = a , lim y = b lim ( f ± g ) = L ± M , lim c f = c L lim f g = L M , lim f g = L M ( M ≠ 0 ) lim f n = L n , lim f n = L n \begin{aligned}\lim c&=c,\quad \lim x=a,\quad \lim y=b\\ \lim(f\pm g)&=L\pm M,\quad \lim cf=cL\\ \lim fg&=LM,\quad \lim\tfrac fg=\tfrac LM\ (M\ne0)\\ \lim f^n&=L^n,\quad \lim\sqrt[n]f=\sqrt[n]L\end{aligned} lim c lim ( f ± g ) lim f g lim f n = c , lim x = a , lim y = b = L ± M , lim c f = c L = L M , lim g f = M L ( M = 0 ) = L n , lim n f = n L
Polynomials: plug in the point.
lim ( x , y ) → ( 1 , 2 ) x 2 y + 1 x + y \displaystyle\lim_{(x,y)\to(1,2)}\frac{x^2y+1}{x+y} ( x , y ) → ( 1 , 2 ) lim x + y x 2 y + 1 → 3 ≠ 0 \to3\ne0 → 3 = 0
= 2 + 1 3 = =\dfrac{2+1}{3}= = 3 2 + 1 = 1 1 1
Same laws as one variable Ask: "Which laws turn a polynomial limit into substitution?" → sum, constant multiple, product, identityRoot law: even n needs L ≥ 0 and f ≥ 0 nearby
Limit of a Function of Two Variables algebra proof · Theorem 4.1
Each law is the one-variable proof with distance in the plane
Split ε, take the smaller δ, use the triangle inequality.
identity law
∣ x − a ∣ ≤ ( x − a ) 2 + ( y − b ) 2 < δ ⇒ δ = ε works |x-a|\le\sqrt{(x-a)^2+(y-b)^2}<\delta\ \Rightarrow\ \delta=\varepsilon\text{ works} ∣ x − a ∣ ≤ ( x − a ) 2 + ( y − b ) 2 < δ ⇒ δ = ε works sum law
∣ f − L ∣ < ε 2 |f-L|<\tfrac\varepsilon2 ∣ f − L ∣ < 2 ε δ 1 \delta_1 δ 1 ∣ g − M ∣ < ε 2 |g-M|<\tfrac\varepsilon2 ∣ g − M ∣ < 2 ε δ 2 \delta_2 δ 2 δ = min ( δ 1 , δ 2 ) \delta=\min(\delta_1,\delta_2) δ = min ( δ 1 , δ 2 )
∣ f + g − ( L + M ) ∣ ≤ ∣ f − L ∣ + ∣ g − M ∣ < ε |f+g-(L+M)|\le|f-L|+|g-M|<\varepsilon ∣ f + g − ( L + M ) ∣ ≤ ∣ f − L ∣ + ∣ g − M ∣ < ε Product and quotient: the same idea with more careful bounds.
Ask: "Why min(δ₁, δ₂)?" → inside both disks at onceBook: "proofs similar to one variable"
Limit of a Function of Two Variables algebra Example 4.8 · Checkpoint 4.6
Check the bottom’s limit first; if it is not 0, substitute
Quotient law needs lim ( bottom ) ≠ 0 \lim(\text{bottom})\ne0 lim ( bottom ) = 0
At ( 2 , − 1 ) (2,-1) ( 2 , − 1 )
(a) x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 x^2-2xy+3y^2-4x+3y-6 x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 = 4 + 4 + 3 − 8 − 3 − 6 = =4+4+3-8-3-6= = 4 + 4 + 3 − 8 − 3 − 6 = − 6 -6 − 6
(b) 2 x + 3 y 4 x − 3 y \dfrac{2x+3y}{4x-3y} 4 x − 3 y 2 x + 3 y → 8 + 3 = 11 ≠ 0 \to8+3=11\ne0 → 8 + 3 = 11 = 0 → 4 − 3 = 1 \to4-3=1 → 4 − 3 = 1 1 11 \dfrac1{11} 11 1
Checkpoint 4.6 lim ( x , y ) → ( 5 , − 2 ) x 2 − y y 2 + x − 1 3 \displaystyle\lim_{(x,y)\to(5,-2)}\sqrt[3]{\frac{x^2-y}{y^2+x-1}} ( x , y ) → ( 5 , − 2 ) lim 3 y 2 + x − 1 x 2 − y
27 / 8 3 = 3 2 \sqrt[3]{27/8}=\tfrac32 3 27/8 = 2 3
Mini-whiteboards before each click Slip: quotient law before checking the bottom CP 4.6: inside 27/8, cube root fine for any sign
Limits That Fail to Exist geometry
In the plane there are infinitely many ways in — the limit must agree on all
Two paths into ( a , b ) (a,b) ( a , b ) lim ( x , y ) → ( a , b ) f \lim_{(x,y)\to(a,b)}f lim ( x , y ) → ( a , b ) f does not exist .
x → a x\to a x → a
( x , y ) → ( a , b ) (x,y)\to(a,b) ( x , y ) → ( a , b )
… or any curve, even a spiral.
Geometry every path ends at the same height
Algebra · p320 substitute the path: one number for all
Ask: "How many ways can x approach 2? (x, y) approach (2, 1)?" → 2 vs infinitely manyPaths can only disprove, not prove
Limits That Fail to Exist bridge Example 4.9a
2 x y 3 x 2 + y 2 \dfrac{2xy}{3x^2+y^2} 3 x 2 + y 2 2 x y y = 0 y=0 y = 0 1 2 \tfrac12 2 1 y = x y=x y = x Find two paths, substitute each, compare: f ( x , 0 ) = 0 f(x,0)=0 f ( x , 0 ) = 0 f ( x , x ) = 1 2 f(x,x)=\tfrac12 f ( x , x ) = 2 1
Along y = 0 y=0 y = 0 0 3 x 2 = 0 \dfrac{0}{3x^2}=0 3 x 2 0 = 0
Along y = x y=x y = x 2 x 2 4 x 2 = 1 2 \dfrac{2x^2}{4x^2}=\tfrac12 4 x 2 2 x 2 = 2 1
ε = 1 4 =\tfrac14 = 4 1 1 2 \tfrac12 2 1
Geometry two walks end at heights 0 and ½
Algebra · p321 f ( x , 0 ) = 0 ≠ 1 2 = f ( x , x ) f(x,0)=0\ne\tfrac12=f(x,x) f ( x , 0 ) = 0 = 2 1 = f ( x , x )
Ask before clicking: "f near the origin?" → most say 0Drag: the graph is a fan of level lines through the z-axis Domain excludes (0, 0)
Limits That Fail to Exist algebra Example 4.9a
Along y = k x y=kx y = k x 2 k 3 + k 2 \dfrac{2k}{3+k^2} 3 + k 2 2 k
Substitute y = k x y=kx y = k x k k k
2 x ⋅ k x 3 x 2 + k 2 x 2 = 2 k 3 + k 2 \dfrac{2x\cdot kx}{3x^2+k^2x^2}=\dfrac{2k}{3+k^2} 3 x 2 + k 2 x 2 2 x ⋅ k x = 3 + k 2 2 k
k = 0 k=0 k = 0 0 0 0
k = 1 k=1 k = 1 1 2 \tfrac12 2 1
Depends on k k k
Do y = kx on the board; x² cancels Ask: "Largest height?" → 1/√3 ≈ 0.58 at k = √3
Limits That Fail to Exist bridge Example 4.9b
Every line gives 0, yet the parabola x = y 2 x=y^2 x = y 2
Agreement along all lines proves nothing; one disagreeing path disproves the limit.
f = 4 x y 2 x 2 + 3 y 4 f=\dfrac{4xy^2}{x^2+3y^4} f = x 2 + 3 y 4 4 x y 2
y = k x y=kx y = k x 4 k 2 x 1 + 3 k 4 x 2 → 0 \dfrac{4k^2x}{1+3k^4x^2}\to0 1 + 3 k 4 x 2 4 k 2 x → 0
x = y 2 x=y^2 x = y 2 4 y 4 y 4 + 3 y 4 = 1 \dfrac{4y^4}{y^4+3y^4}=1 y 4 + 3 y 4 4 y 4 = 1
Geometry lines miss the ridge on
x = y 2 x=y^2 x = y 2 Algebra · p322 path limits
0 ≠ 1 0\ne1 0 = 1 ⇒ DNE
Ask: "k = 1, 2, 10 all give 0 — limit 0?" → most say yesSlip: "two paths agree ⇒ limit exists" Why lines miss: Level 3
Limits That Fail to Exist algebra Checkpoint 4.7
Along lines through ( 2 , 1 ) (2,1) ( 2 , 1 ) k 1 + k 2 \dfrac{k}{1+k^2} 1 + k 2 k
At ( a , b ) ≠ ( 0 , 0 ) (a,b)\ne(0,0) ( a , b ) = ( 0 , 0 ) ( a , b ) (a,b) ( a , b ) y − b = k ( x − a ) y-b=k(x-a) y − b = k ( x − a )
Checkpoint 4.7 lim ( x , y ) → ( 2 , 1 ) ( x − 2 ) ( y − 1 ) ( x − 2 ) 2 + ( y − 1 ) 2 \displaystyle\lim_{(x,y)\to(2,1)}\frac{(x-2)(y-1)}{(x-2)^2+(y-1)^2} ( x , y ) → ( 2 , 1 ) lim ( x − 2 ) 2 + ( y − 1 ) 2 ( x − 2 ) ( y − 1 )
h = x − 2 h=x-2 h = x − 2 k h 2 h 2 + k 2 h 2 = k 1 + k 2 \dfrac{kh^2}{h^2+k^2h^2}=\dfrac{k}{1+k^2} h 2 + k 2 h 2 k h 2 = 1 + k 2 k
k = 0 k=0 k = 0 k = 1 \ k=1 k = 1 1 2 \tfrac12 2 1
Depends on k k k
Pairs, 3 min Same shape as Example 4.9a, moved to (2, 1)
Limits That Fail to Exist algebra Example 4.9b · why
Every line near the origin slides onto flat parabolas, so it misses the ridge
On x = k y 2 x=ky^2 x = k y 2 f = 4 k k 2 + 3 f=\dfrac{4k}{k^2+3} f = k 2 + 3 4 k y = m x y=mx y = m x k = 1 m 2 x → ∞ k=\dfrac1{m^2x}\to\infty k = m 2 x 1 → ∞ f → 0 f\to0 f → 0
k = 1 k=1 k = 1 2 / 3 ≈ 1.15 2/\sqrt3\approx1.15 2/ 3 ≈ 1.15 k = 3 k=\sqrt3 k = 3
A line: k = 1 m 2 x → ∞ k=\dfrac1{m^2x}\to\infty k = m 2 x 1 → ∞ x → 0 x\to0 x → 0
… where the height goes to 0.
Check: 4k/(k² + 3) at k = 1/(m²x) equals 4m²x/(1 + 3m⁴x²)
Interior Points and Boundary Points geometry Figure 4.17
A point is interior if some disk fits inside S S S
Interior : some δ disk around P 0 P_0 P 0 S S S Boundary : every δ disk around P 0 P_0 P 0 S S S
( − 1 , 1 ) (-1,1) ( − 1 , 1 )
( 2 , 3 ) (2,3) ( 2 , 3 )
Geometry disk inside vs disk across the edge
Algebra · p322 some δ: disk
⊂ S \subset S ⊂ S · every δ: in and out
Ask: "However small the disk at (2, 3)?" → still crossesBoundary points need not belong to S
Interior Points and Boundary Points bridge Definitions
Open leaves its edge out, closed keeps all of it — half an edge is neither
Open : every point interior. Closed : contains all boundary points. Region : open, connected, nonempty.
δ disk, edge left out: open .
Edge included: closed .
Half the edge: neither.
Geometry dashed edge · solid edge
Algebra · p323 ⋯ < δ 2 \ \cdots<\delta^2 ⋯ < δ 2 ·
⋯ ≤ δ 2 \ \cdots\le\delta^2 ⋯ ≤ δ 2
Ask: "The whole plane: open or closed?" → bothConnected: not two separate open pieces
Interior Points and Boundary Points bridge Definition · Example 4.10
At a boundary point of the domain, approach only through the domain
Same ε–δ definition, using only points in both the δ disk and the domain. The limit laws still apply.
lim ( x , y ) → ( 4 , 3 ) 25 − x 2 − y 2 \displaystyle\lim_{(x,y)\to(4,3)}\sqrt{25-x^2-y^2} ( x , y ) → ( 4 , 3 ) lim 25 − x 2 − y 2
Domain x 2 + y 2 ≤ 25 x^2+y^2\le25 x 2 + y 2 ≤ 25 ( 4 , 3 ) (4,3) ( 4 , 3 )
Come in from inside: 25 − 16 − 9 = \sqrt{25-16-9}= 25 − 16 − 9 = 0 0 0
Geometry half of every disk lies outside
Algebra · p323 ε–δ with
( x , y ) (x,y) ( x , y ) in the domain
Ask: "Which half of the disk around (4, 3) counts?" → the part inside x² + y² ≤ 25LO 4.2.2: boundary points and paths
Interior Points and Boundary Points algebra Checkpoint 4.8
29 − x 2 − y 2 \sqrt{29-x^2-y^2} 29 − x 2 − y 2 ( 5 , − 2 ) (5,-2) ( 5 , − 2 ) 5 2 + ( − 2 ) 2 = 29 5^2+(-2)^2=29 5 2 + ( − 2 ) 2 = 29
Checkpoint 4.8 lim ( x , y ) → ( 5 , − 2 ) 29 − x 2 − y 2 \displaystyle\lim_{(x,y)\to(5,-2)}\sqrt{29-x^2-y^2} ( x , y ) → ( 5 , − 2 ) lim 29 − x 2 − y 2
29 − 25 − 4 = 0 \sqrt{29-25-4}=0 29 − 25 − 4 = 0
Domain: x 2 + y 2 ≤ 29 x^2+y^2\le29 x 2 + y 2 ≤ 29 29 \sqrt{29} 29
Pairs, 2 min Same as Example 4.10 with radius √29
Continuity of Functions of Two Variables bridge Definition
Continuous at ( a , b ) (a,b) ( a , b )
1. f ( a , b ) f(a,b) f ( a , b ) lim ( x , y ) → ( a , b ) f \lim_{(x,y)\to(a,b)}f lim ( x , y ) → ( a , b ) f lim ( x , y ) → ( a , b ) f = f ( a , b ) \lim_{(x,y)\to(a,b)}f=f(a,b) lim ( x , y ) → ( a , b ) f = f ( a , b )
f = sin ( x 2 + y 2 ) x 2 + y 2 f=\dfrac{\sin(x^2+y^2)}{x^2+y^2} f = x 2 + y 2 sin ( x 2 + y 2 ) ( 0 , 0 ) (0,0) ( 0 , 0 )
1. no value at ( 0 , 0 ) (0,0) ( 0 , 0 )
2. sin s s → 1 \dfrac{\sin s}{s}\to1 s sin s → 1
3. define f ( 0 , 0 ) = 1 f(0,0)=1 f ( 0 , 0 ) = 1
Geometry no hole, no tear over
( a , b ) (a,b) ( a , b ) Algebra · p325 lim f = f ( a , b ) \lim f=f(a,b) lim f = f ( a , b )
Ask at each click: "Does the bump pass?"Slip: "0/0 ⇒ no limit" — here 0/0 → 1 Contrast: (x² − y²)/(x² + y²) fails 2; no value repairs it
Continuity of Functions of Two Variables algebra Example 4.11 · Checkpoint 4.9
Verify continuity by checking all three conditions at the point
Value, limit, equal — write all three, even when substitution makes them look obvious.
f = 3 x + 2 y x + y + 1 f=\dfrac{3x+2y}{x+y+1} f = x + y + 1 3 x + 2 y ( 5 , − 3 ) (5,-3) ( 5 , − 3 )
f ( 5 , − 3 ) = 15 − 6 5 − 3 + 1 = 3 f(5,-3)=\dfrac{15-6}{5-3+1}=3 f ( 5 , − 3 ) = 5 − 3 + 1 15 − 6 = 3 bottom → 3 ≠ 0 \to3\ne0 → 3 = 0 = 3 =3 = 3 3 = 3 3=3 3 = 3 continuous
Checkpoint 4.9 26 − 2 x 2 − y 2 \sqrt{26-2x^2-y^2} 26 − 2 x 2 − y 2 ( 2 , − 3 ) (2,-3) ( 2 , − 3 )
26 − 8 − 9 = 3 \sqrt{26-8-9}=3 26 − 8 − 9 = 3
Book key p861 prints 16 for 26 — its 16 − 8 − 9 16-8-9 16 − 8 − 9
Ask: "Where is f not continuous?" → on x + y + 1 = 0CP 4.9 erratum: key p861 prints √(16 − 2(2)² − (−3)²) = 3; correct is 26
Continuity of Functions of Two Variables algebra proof · Theorem 4.2
In ε–δ form, continuity lets L L L f ( a , b ) f(a,b) f ( a , b )
Continuous at ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) ∣ f ( x , y ) − f ( x 0 , y 0 ) ∣ < ε |f(x,y)-f(x_0,y_0)|<\varepsilon ∣ f ( x , y ) − f ( x 0 , y 0 ) ∣ < ε ( x − x 0 ) 2 + ( y − y 0 ) 2 < δ \sqrt{(x-x_0)^2+(y-y_0)^2}<\delta ( x − x 0 ) 2 + ( y − y 0 ) 2 < δ Theorem 4.2: f f f g g g ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) δ = min ( δ f , δ g ) \delta=\min(\delta_f,\delta_g) δ = min ( δ f , δ g ) ε 2 \tfrac\varepsilon2 2 ε
∣ ( f + g ) − ( f + g ) ( x 0 , y 0 ) ∣ ≤ ∣ f − f ( x 0 , y 0 ) ∣ + ∣ g − g ( x 0 , y 0 ) ∣ < ε |(f+g)-(f+g)(x_0,y_0)|\le|f-f(x_0,y_0)|+|g-g(x_0,y_0)|<\varepsilon ∣ ( f + g ) − ( f + g ) ( x 0 , y 0 ) ∣ ≤ ∣ f − f ( x 0 , y 0 ) ∣ + ∣ g − g ( x 0 , y 0 ) ∣ < ε The sum-law proof, with L L L M M M
No "0 <" here: the point itself is allowed Theorem 4.3 (g(x)h(y)) by the product bound
Continuity of Functions of Two Variables bridge Theorems 4.2–4.4 · Figure 4.20
Sums, products and compositions of continuous functions are continuous
f + g f+g f + g g ( x ) h ( y ) g(x)h(y) g ( x ) h ( y ) f ∘ g f\circ g f ∘ g
g g g ( x , y ) ↦ z (x,y)\mapsto z ( x , y ) ↦ z
f f f z ↦ f ( z ) z\mapsto f(z) z ↦ f ( z )
Small disk → short interval → short interval.
Geometry close inputs stay close at every step
Algebra · p326 f f f cont. at
g ( x 0 , y 0 ) g(x_0,y_0) g ( x 0 , y 0 ) ⇒
f ∘ g f\circ g f ∘ g
Ask: "Where can a formula built this way fail?" → zero bottom, log of ≤ 0, even root of < 0Theorem 4.3 needs g(x) and h(y) separately continuous
Continuity of Functions of Two Variables algebra proof · Theorem 4.4
Composition: chain the two ε–δ promises
ε for f f f z 0 z_0 z 0 g g g f f f z 0 = g ( x 0 , y 0 ) z_0=g(x_0,y_0) z 0 = g ( x 0 , y 0 ) ∣ z − z 0 ∣ < η ⇒ ∣ f ( z ) − f ( z 0 ) ∣ < ε |z-z_0|<\eta\Rightarrow|f(z)-f(z_0)|<\varepsilon ∣ z − z 0 ∣ < η ⇒ ∣ f ( z ) − f ( z 0 ) ∣ < ε
g g g ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) dist < δ ⇒ ∣ g ( x , y ) − z 0 ∣ < η \text{dist}<\delta\Rightarrow|g(x,y)-z_0|<\eta dist < δ ⇒ ∣ g ( x , y ) − z 0 ∣ < η
dist < δ ⇒ ∣ f ( g ( x , y ) ) − f ( g ( x 0 , y 0 ) ) ∣ < ε \text{dist}<\delta\ \Rightarrow\ |f(g(x,y))-f(g(x_0,y_0))|<\varepsilon dist < δ ⇒ ∣ f ( g ( x , y )) − f ( g ( x 0 , y 0 )) ∣ < ε Ask: "Which δ do we find first?" → η for f, then δ for gFigure 4.20 is this chain
Continuity of Functions of Two Variables algebra Example 4.12 · Checkpoint 4.10
Name the continuous pieces and the theorem that joins them
Product: g ( x ) h ( y ) g(x)h(y) g ( x ) h ( y )
4 x 3 4x^3 4 x 3 y 2 y^2 y 2 4 x 3 y 2 4x^3y^2 4 x 3 y 2
cos \cos cos cos ( 4 x 3 y 2 ) \cos(4x^3y^2) cos ( 4 x 3 y 2 ) everywhere
Checkpoint 4.10 2 x 2 y 3 + 3 2x^2y^3+3 2 x 2 y 3 + 3 ( 2 x 2 y 3 + 3 ) 4 (2x^2y^3+3)^4 ( 2 x 2 y 3 + 3 ) 4
2 x 2 ⋅ y 3 2x^2\cdot y^3 2 x 2 ⋅ y 3 + 3 +\,3 + 3 u 4 u^4 u 4
Constant function continuous everywhere Ask: "Is ln(4x³y²) continuous everywhere?" → no, only where 4x³y² > 0
Functions of Three or More Variables bridge Definition · Example 4.13
In three variables the δ disk becomes a δ ball — the limit laws are unchanged
δ ball : ( x − x 0 ) 2 + ( y − y 0 ) 2 + ( z − z 0 ) 2 < δ \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}<\delta ( x − x 0 ) 2 + ( y − y 0 ) 2 + ( z − z 0 ) 2 < δ
Smaller δ, smaller ball around ( 4 , 1 , − 3 ) (4,1,-3) ( 4 , 1 , − 3 )
lim x 2 y − 3 z 2 x + 5 y − z \displaystyle\lim\frac{x^2y-3z}{2x+5y-z} lim 2 x + 5 y − z x 2 y − 3 z → 16 ≠ 0 \to16\ne0 → 16 = 0 → 25 \to25 → 25 25 16 \dfrac{25}{16} 16 25
Geometry solid ball of radius δ
Algebra · p327 ⋯ + ( z − z 0 ) 2 < δ \sqrt{\cdots+(z-z_0)^2}<\delta ⋯ + ( z − z 0 ) 2 < δ
Ask: "A δ ball in ℝ⁴?" → add (w − w₀)²Continuity in 3D: same three conditions
Functions of Three or More Variables algebra Example 4.13 · Checkpoint 4.11
Three variables, same routine: bottom first, then top, then divide
Check lim ( bottom ) ≠ 0 \lim(\text{bottom})\ne0 lim ( bottom ) = 0 ( x 0 , y 0 , z 0 ) (x_0,y_0,z_0) ( x 0 , y 0 , z 0 )
lim ( x , y , z ) → ( 4 , 1 , − 3 ) x 2 y − 3 z 2 x + 5 y − z \displaystyle\lim_{(x,y,z)\to(4,1,-3)}\frac{x^2y-3z}{2x+5y-z} ( x , y , z ) → ( 4 , 1 , − 3 ) lim 2 x + 5 y − z x 2 y − 3 z
bottom: 2 ( 4 ) + 5 ( 1 ) − ( − 3 ) = 16 2(4)+5(1)-(-3)=16 2 ( 4 ) + 5 ( 1 ) − ( − 3 ) = 16
top: ( 4 2 ) ( 1 ) − 3 ( − 3 ) = 25 (4^2)(1)-3(-3)=25 ( 4 2 ) ( 1 ) − 3 ( − 3 ) = 25
limit 25 16 \dfrac{25}{16} 16 25
Checkpoint 4.11 lim ( x , y , z ) → ( 4 , − 1 , 3 ) 13 − x 2 − 2 y 2 + z 2 \displaystyle\lim_{(x,y,z)\to(4,-1,3)}\sqrt{13-x^2-2y^2+z^2} ( x , y , z ) → ( 4 , − 1 , 3 ) lim 13 − x 2 − 2 y 2 + z 2
13 − 16 − 2 + 9 = 4 = 2 \sqrt{13-16-2+9}=\sqrt4=2 13 − 16 − 2 + 9 = 4 = 2
Mini-whiteboards Slip: sign of −(−3) in the bottom
Section 4.2 Exercises algebra Exercises 95, 107
Polar coordinates: a bound in r r r
x = r cos θ , y = r sin θ x=r\cos\theta,\ y=r\sin\theta x = r cos θ , y = r sin θ ( x , y ) → ( 0 , 0 ) (x,y)\to(0,0) ( x , y ) → ( 0 , 0 ) r → 0 r\to0 r → 0 2 x y 3 x 2 + y 2 = 2 cos θ sin θ 3 cos 2 θ + sin 2 θ no r : no limit \frac{2xy}{3x^2+y^2}=\frac{2\cos\theta\sin\theta}{3\cos^2\theta+\sin^2\theta}\quad\text{no }r\text{: no limit} 3 x 2 + y 2 2 x y = 3 cos 2 θ + sin 2 θ 2 cos θ sin θ no r : no limit ∣ x 2 y x 2 + y 2 ∣ = r cos 2 θ ∣ sin θ ∣ ≤ r → 0 \Big|\frac{x^2y}{x^2+y^2}\Big|=r\cos^2\theta\,|\sin\theta|\le r\to0 x 2 + y 2 x 2 y = r cos 2 θ ∣ sin θ ∣ ≤ r → 0 sin x 2 + y 2 x 2 + y 2 = sin r r → 1 \frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}=\frac{\sin r}{r}\to1 x 2 + y 2 sin x 2 + y 2 = r sin r → 1 Line 2 = Exercise 95 continuous with f(0, 0) = 0 Line 3 = Exercise 107
§4.2 wrap-up
A limit in the plane must agree along every path; continuity is limit = value
Laws to compute · two paths to disprove · three conditions to verify continuity.
Objective You can now
4.2.1 limit in two variablescheck the bottom, then substitute with Theorem 4.1 4.2.2 paths and boundary pointstwo paths with different limits ⇒ DNE; at the edge use the domain only 4.2.3 conditions for continuityvalue exists · limit exists · equal 4.2.4 verify continuity at a pointExample 4.11; build from pieces by Theorems 4.2–4.4 4.2.5 three or more variablesδ ball; same laws: Example 4.13 → 25/16
Exit ticket: (x² − y²)/(x² + y²) at (0, 0) → x-axis 1, y-axis −1, DNE Homework: 61, 63, 65, 69, 73, 77, 81, 83, 85, 87, 89, 91, 93, 95, 97, 101, 107, 109