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LASTTIME
Limite di funtionsin pin variabili
7:IR"-IR
Riconola: ommette in paR" limite L ->(R,
tim
f(x) L S =
1- (
=
1 aI · L 5
+
seV2x75x0:( -
1)(8 =
(f(x) - 2)<5
07/03/2023 Analisi Matematica 2 9 CFU
Esempir
5xxy S y gn,
(,y) (0,07 (,y) (0,07 =
z
=
7(x,y) =
g(x,y) =
(x,y) (0,0)
=
(x,y) (0,0) =
Pa
quake fra f, esist is limite
him
him
f(x,y), g(x,y)
(x,y) (0,0)
->
(x,y) (0,0)
-
DNE
Joked:coondinate
polan x xo fc0s8
+
=
f/0
y-..... (x,y)
[0,27)
-
y 90 f20
↓
=
+
0 ->
f y 928
=
10 (x,y) -
(7,0)
(X0,90X 9208
=
((x(1,0),y(9,0)) -
(X0,y )
-
=
T
5x75>0:
V
lin f(x,y):
(x,y) -
x(X0,y0) (f(x,y)) =5 1(f(X,y1
= -
L) x9
Aconguo
f(x,y) =
1yz
=
=e0 vino
h
=
72
x0 0f(x,y)
=
0
=
in (f(x,21-0) 100201 *
general
0
=
1908Kind
g(x,y)
myzy ff0
=
=
=
(8(X,y) -
0) 1/
=
1 1
=
< s
2
peewlo (f1 x =
8
sup
Abbiano maggiona
*Otto, in
(g(X0 7218,90
+
g20)
+
-
L) h(f) =
can tif) to
press fro
Terramat[criteria del
shiesistent limite]
X
x0
= +
y200
Consideriama lin
f(x,y) ->
(x,y) +
70,70) y 7
=
ynd
+
Se esiste it 1R
unafactions h:(0,0) [0,0)
->
per ogm
8 + [0,37] (f(x0 92050, 30 72n8)
+ + -
L) h(e)
<
1 x 0
allow lin
f(x,y) L
=
(x,y) 70,70)
+
Dim.
Escitio teavare lim x209y ->
-
o
Verifical (X,y) -
(0,1) (X' (y
+
- 1(3)"2
V010,11
(0,y)-(0,4))dy,,,
·
lin 0
=
x0
N
X
v
=
t
x 0
f20s8
= +
log (1 t)
+
y =1 +
ysin0
- =
⑭
120501 Roy (1 sina) /
t
+
=
<
1801 lysing
t) t
log (1 +
=
7 Icowning
I
- (1 + 1 27
1 2
=
+
=
·
<
T ->
7-0
Riconde f(x,y) continue
I
in (0,70) se lim f(x,y) f(x0,y0)
=
(x,y) -
3(X0,y0)
Esempie sia x3 y2
S
+
+
(x,y)f(0,0)
f.(R- - (R. f(x,y) x
=
yz
0 (x,y) (0,0)
=
Trovara talti
partin cuif
i a continue.
Esempio
Studiara lacontinuita
della funtions (50,70)
7
continue in
ex/y -
y =
0sewhx700.
X
f(x,y) =
S 0
y 0
=
a. (50,90) con
g0 = 0 continue
b.(X0,80),X0=0,y0 0
=
Esempio
Studiara lacontinuita
della funtions (50,70)
7
continue in
ex/y + 0/
y sewhx400.
X
f(x,y) =
S 0
y 0
=
a. (50,90) con
g0 = 0 continue
a
X x.
=
+
t
V y
=
b.(X0,80),X0=0,y0 0
=
2x0
-
him x0 40,7 + 0
= =
t -> 0I t -> st -
X0>0,t+0
existe -o+
non
x0x0,t
C.(X0,y0) (0,0) =
+
+
relte x
t,y
=
= t tim fCt,t) lim
=
+
1 =0
+3>0I +30
1st
H 1t
t,y t2 tim f(t,th) him te him e
anwax
=
= = =
=
X/t
+
+->* +->00 ⑳ + -
H
=
wine" ( -
((s) 0
=
+
- 0
(((t")
non exite
Eseccitio
y28(1 xy)
3
+
xyx
-
1,y 0
+
f(x,y) =
0 eltrimenti
Studiare
pantislicontinuita
i dif
·
tone rende fcontinue.
1516)
:
⑥
on biance to contine
auwe
y 7 ne
=
.
OK s
a
+ &
8A · continue
·minine
is
it ·assey
Si:
⑥
·asseX,X>0 No
Asse
y:lin f(0,y) 0
=
y + y0
f(x,y) (+jcosae)
colg
Sniviamo x 720s0
= =
y 90 ffin&
= +
0x - M(2,M/]
f(x,y) (+jcosae)
colg 40
= =
0
=
(1990) y)
< -> 0
Cin0 0
=
7 - 0
i
manca no penti (x0,0) x 10
=
98010
+
y 0
=
fua
+
f(x,y) =
log( xy)
+
(x0 1910)
=
+
fsiu d
[easi0 an e
x720
= xo
+
2
2
- C*
7250)(0)7
0 = x0(X0 9200)
+
= xo
-
sin Q
⑤ -
-
Differentiabilita (n 1) =
f(x0) a(x x)-
+ -
Af(x,x0)
=
f(x0)
·
Lareta
y
= a(X x))
+ -
f(x0) -
f(x)
i
- - - . . .
a
tangente at
prefico
↓finX0s
him Af(x,x0) 0
=
x0 x
x X0 x -
x0
-
2 0= ⑦
2,,
=
f(x) -
f(X-) -
- f'(x0)
x - x0
Ax f(xa Ax
Af(X,x-) f
=
f(+ - a
↓
x -
x0
↓ ↓
f(t-f(x)
0 an 9
=
x -1 x0
Differentiabilita (n 1) =
f(x0) a(x x)-
+ -
Af(x,x0)
=
f(x0)
·
Lareta
y
= a(X x))
+ -
f(x0) ·
f(x)
:
- - - - - - - -
a
tangente at
prefico
↓finX0s
!
him Af(x,x0) =
0 (A) x⑤ x
X X0
-
x -
x0
[f(x-) a(x
+ -
x0]
x-
X0 x -
xy
X +
X0 x -
xo
In
questo caso dciam chefdifferentiable in to
Piano
tangenta
Sia f:AC(R" 1R
->e
p (p1,..., Pn)
=
- A
I
generico piano passante (p,,..., Pn,f(ph) in IR**:
pen
t -
f(p1,..Pu) 01(X,
=
-
p,) +... Qu(xu -
pu)
t f(p) =
a.(X -
1)
&adefinit
-
tim f(x) -
[f(p) a(1
+
-
11] 0
↑ affinchi
cosi
is
=
12 pier sis
x
-p
-
1/
tangente.
-t f(x,xz)
=
⑧
1
p.) P2)
(x1,X2,t):t f(p) a,(x, a.(xc -
+
=
- -
↑
10y,a2, -
1)
X2
/
↓ p (P.,P2)
=
X,
Piano
tangenta
Sian f: Ac(R" -1Re p (p1,..., Pn)
=
- A
I
generico piano passante (p,,..., Pn,f(ph) in IR**:
pen
(x,,...,Xn,t):t -
f(p) a1(X,
= -
P1) ..
+
an(X
+ -
Pn)
=
a -
( -
p)
Conditiona
tengeuta:
tim f(x) -
[f(p) 2
+
(x- p)]
0
=
x
- |X p)
p
-
-t f(x,xz)
=
⑧
1
p.) P2)
(x1,X2,t):t f(p) a,(x, a.(xc -
+
=
- -
↑
10y,a2, -
1)
X2
↓
/ p (P.,P2)
=
X,
ACIRY 1
Bef: differentiable in
i
-
>
sa
a
per 0
=
x ->
p 1x -
Teorema ↑
2)Sefi differentiable in
1, allowe fcontinuein
b) se fi siferentichile in
I
alone a (0, .
=
. .
., aj)
e do
olato
+ -> 0 t
=
2xjf(p) 2jf(I) =
·(P.,Pz t)+
⑦ Vf(p)
= =
(21f(p),...Gjf(p),..) ·
P1,P)
Bf:ACIR" -> R
I differentiable in
p se
a
per 0
=
x ->
p 1x -
Teorema ↑
a) Sef differentiable
a in
1, allow tocontinuein
b) Set a differentiable in
I,
allows is rettone a (0,..., an)
=
soolsh's fa
F(P1, f(P)
aj
lin
=
t 0
->
..., pj +t,... pu)
t
-
2x;f(x 1,...,Xn))x 1
=
ie
& secondo member a laderivate
particle of simpette
ellavariabila
xj
grediente
Dim if differentiable in
p=> of continuou
[f(p) a.(X 1)
1:7: f(x)
him
f siff. In
- + -
0
=
x
p
->
(x -
17
==> Vp>74570s
(f(x) [f(p) a.(x
8 -
p)])
=
=>
B
+ -
-
(x -
pl
=> (f(x) -
f(p) -
a(x -
1)/ <B1x -
31
== (f(x) -
f(p)) p(X =
-
p) +
18 .(X 1)) -
= B(X 1) - +
18)(X -
1)
CS
B((x)
=
+
p)x @
scytiousp <min
35,03
Unol hre (x- p) < 8 =) (f(x) -
f(p)) < I
i- we
him f(P1...pi + t...pu) -
+
- x
t
tej
-
f(1 tej)
+ -
t-0 |tej)
-
-
~
⑦ .
ej
7f(p)
-
tin
a
↑
+zot
~
)
·
rex
=
#aj aj
=
lin f(x,y) - +
⑧
[f(X0,401 Vf(X0,40).(X x0,4 40)]
- -
-
0
t(0)
=
tv
- Nox,y)
V (070) 75
= +
-
↓
>3 55
05]
M
↑
ein f(*0,301 w)
+ -
[(X0,30) +*
t -0 tIv/
f(x)
0
=
3
=
+00 aifat)
- Wf(x0,40 =
(xf(X0,90) hinf(X0=
t,y0)
+
-
f(X0,y0)
+- 0 t
by f(X0,90) =
ein f(X0,40 +1
+ -
f(X0,4)
+ 10
t
lin f(x) -
[f(p) a.(X
+ -
)] 0
=
x -
p |x -
p)
cide p
X x 07 S (0,1)
+
|X p(XS
-
1(f(x)
= -
f(p) a(x
+ -
p)) +
B
|x -
p|
p1 +
1a/B <
- B(S (a)) p(1
+
< 1a))
+
Felto!
b) Usanolo f(X) -
f(p) 1)(x + -
0
=
+->0
t
+ 10 t
directionalSia
Derivate v (v1,
=
..., W.) un
qrdsiasiwettors
shi154
limite
f(f)
+ 1> 0 ↳
Adesempio se
ej (1,...
=
5,...0)
linf(P1,...pi f(P)
Pojf(p) 1....Pu)
ajf(p)
-
+
=
=
t -> 0 -
Syf(p) =
Vf(p) .w
Riassumenoo:TGCP)
(Zf(p),..., f(p)
=
-> GRADIENTE
f differentiabilein so
a a solo se him f(1+2) -
[f(p) h
+
.vf(p)] 0
=
a - 0 |h|
her
qual caso Dvf(p) Vf(p).5 =
directionalSia
Derivate v (v1,
=
..., W.) un
qrdsiasiwettors
shi154
limite
Duf(p) =
Teorema 1x
f(x,y) ly(x yyz)
=
+
+
e ouH.in (40,401 =(0,0)
calcobe VE(X0,70)
2xf(X0,40) him
=
f(X0 t,y0)
+
-
f(x0,40)
+ ->0 ↳
=
him log((x0 1)
+
(0
+ +
30] -
fy(x0 2y)
+
+ a0
t
~
g((0) h(x0)
=
=
love
g(t) =
Cog(0 7) (x
+ + +
40")
h(x) 2y,3)
20y(x
=
+
I ↑
h(x +t1 -
h(X!
u'(x,
[ vx
=
y8
+
(1
+
=y,)Ix
1
x-
=
2xf(x,y) =
x
x y
(1
Eiy)
+
byf(x,y) =
*
(xi yu (
iyz)
2 +
+
ES: f(x,y)
xy/2xf(x,y)
=
2xy
=
+ ->0 t t
x y 2xxy 4e xy
y
+
+ -
2xy
= +
7yJ 2xy
=
Esempio
Dimostre cha f(x,y) 2x
= x y
i differentiabile in (0,0)
⑦ = e 22
=
- x =
0
7 =
x
+ x3 yz
2yf(x,y) 2y[2xe
x
y)
↓
+
-
4xy
=
e
=
differntiobilite: [2,07 y 0
=
Verifica
-/
him ⑫ (71 - +
(2,07.[x,y]
[f(0,0) Vf(0,0)[(x,y) (0,01] -
(X,y(-(0,0) 1(x,y) -
10,01/
x+ yz
1
-
him
[e
2x -
=
0
(x,y) (0,0)
[X' y-) /2
-
+
g(x,y)
[eT(sina
2010)
1/
-
127800
18(ye0s0,72u0)1
-
f
[eT"(sine
2010)
1]/
-
127800
18(yes0,720))
-
f
uso asintotic et t
(et -
1) 27
=
2
=2 1850) f (sin0 -
2010)
- -
= 1 =2
2
< 49 -05710
Esempio
x-y i
f(x,y)
-
t
linf(t,0 7(0,0)
-
2t
zf(0,0) lin
=
-
e
=>
-
- 0
t + 10 t
=
1
f(0,0) =>
lin
+ 10
f(0,
t1- f(0,0) 0
=
(u,v).(2,0)
Si haaxlesspa z (u,5)
=
(u,5)-(0,0 1(u,v))
-
u 02
+
lim
=
2ue - 2u
(n+12)
him
=
2u -
e - 4] =
lim -
2u(u- v2) +
v2/z
(u2 v3)"2
(u,0) - 10
+ ↓ (u,v)-0(uz +
lin.notevok
0!
=
eft - 1
Eampia Si simostri che
f(x,y) x yz
&
+
(0,0)
(x,y)
=
a continuain
·
10,01
proguid
/
↓
·
Non sufferentiable in (0,0)
lin f(x,y) 0
=
+(010, ri8)
!
(x,y) ->
(0,0)
27 2505,7
1
If(9000, 1001
1 (800) 11an8
=
=
pe 10
flepsin
aidtyf
=
01
+
=- 1
<
1 +8
as
+ 18
-
-
0
Peoso, f
(0,0) lin
=
f(tes0, 5an8) -
f(0,01
and
=10
lin
=
E(,0)'sn8
+20 t
os
=
"
Wf(0,01 [Perso,vino, :
Beach,an4x] [0,0] =
Esacizio
+ -> 0 t
f(x0,y0 f(x0,y0)
Eyf(X0,y0) him
= t1
+ -
t -> 0
t
di1 R la funtions
in
ognipento per
(x,y) (0,0)
9
5
f(x,y)
=
e si dice sedifferentiable in
(0,00 0,0) =
Solutions
m
Esacizio
+ -> 0 t
f(x0,y0 f(x0,y0)
Eyf(X0,y0)
=
him t1
+ -
t -> 0
t
di1 R la funtions
in
ognipento per
(x,y) (0,0)
9
5
f(x,y)
=
e si dice sedifferentiable in
(0,00 0,0) =
Solutions
m
=
-
xy.2x -
(xz y2)z+
Se (x,y) (0,0)
=
+->5
f(0,t) f(0,0
2f(0,9
-
= lim - - 0
-
- 10 t
in
sooffosse sufferentiable 10,01 si arable
lin f(x,y) -
[f(0,0) -
2f(0,0).X -
27(0,03] -
-
O
tutteriaquesto
equivale a
lim xy
⑰
(x,y) >(0,0) (x- y-(3/2
+
scegliendo lacanva x =
y, tultavia
him
x+ 0
[ (x2 y23x)
x y
x
=
21 x
x2
13
7 =
+ 0
y
+
limite in # existe
quindsit non
Donena:eoso
m possions
dire a
glialtarpanti (x,y)
in
adifferentiabilita