数学分析(复旦大学版)课后题答案1-6

r(Mn, M0) < ε
n→∞
= (xn − x0)2 + (yn − y0)2 < ε u´½k|xn − x0| r(Mn, M0) < ε, |yn − y0|
= r(Mn, M0) < ε xn → x0, yn → y0(n → ∞)
⇐
Ï = (|xn − x0| + |yn − y0|)2 |xn − x0|2 + |yn − y0|2 0
(5) ½Â÷vØ1ªr2 x2 + y2 + z2 R2'X8
2. ¦e4µ
x2 + y2 (1) lim
x→0 |x| + |y|
y→0
(2) lim
x→0 y→0
x2 + y2 x2 + y2 + 1 − 1
1 + x2 + y2 (3) lim
x→0 x2 + y2
y→0
sin(x3 + y3)
204
1o õCþÈ©Æ
1Ü© õ£¼ê4Ø
1nÙ õ£¼ê4ëY
§1. ²¡:8
y² '¿^´µ 1.
(xn, yn) → (x0, y0)
xn → x0, yn → y0(n → ∞)
y²µ⇒
Ï §ué §¨ §k lim Mn = M0
∀ε > 0, ∃N ∈ Z+ n > N
x→0
y→0
q ØQ ¨ § Ø 1
1
lim (x + y) · sin · sin
1
1
1
x = (k = ±1, ±2, · · · ) lim (x + y) · sin · sin
y→0
x
y
kπ
x→0
x
y
Q¨ 1 y = (k = ±1, ±2, · · · )
kπ
= 9 ÑØQ lim lim f(x, y) lim lim f(x, y)
6. yy²²²µ¡⇒X'Âñ¦n.
# §ué §¨ §k Mn → M0(n → ∞)
∀ε > 0, ∃N ∈ Z+ n, m > N
ε
ε
r(Mn, M0) < 2 , r(Mm, M0) < 2
205
dål'nØ1ª§&r(Mm, Mn) r(Mn, M0) + r(Mm, M0) < ε
3. 4.
yyV¦²²eµµKeÏX=M8MNnnE§'→ →uSMMéX00((§nn©→ →kX>∞ ∞§K))§§b§@uFkéoXn§∀kµε>'>n?0K,Û∃=Nn∈NfZ +N§M§¨ngkn →,> kNMr0(.M§nkk ,rM(M0)n
= , M0) < ε
< ε Mnk
(5) u R2 − x2 − y2 − z2 + x2 + y2 + z2 − r2
)µ
(1) ½Âx 0 y 1
(2) ½Â÷vØ1ªy x + 1'X8
(3) ½Â²¡x + y < 0
½Â÷vØ1ª 'X8 (4)
2kπ x2 + y2 (2k + 1)π(k = 0, 1, 2, · · · )

=
u´ k O(Mi, δMi)õ¹E¥kX i=1
¢du §u´gñ k
O(Mi, δMi ) ⊃ R ⊃ E
.
i=1
206
§2.
1. (½¿±Ñe¼ê½Âµ √√ (1) u = x − 1 − y √ (2) u = x − y + 1
õ£¼ê4ÚëS
(3) u = ln(−x − y) (4) u sin(x2 + y2)
=
1 n
§QO(M0,
δ1)¥½QE'XM1
=
¶Q M0 O(M0,
δ2)¥½
Xu´d¨cIe§&§ X = n → ∞
1 {Mn}(Mn = Mi)(i = 0, 1, · · · , n − 1) r(M0, Mn) < n r(M0, Mn) → 0 Mn → M0(n → ∞).
x→0
1 x
.
~µ
1
(3)
f (x, y) =
x sin y
,
y = 0 QX(0, 0)'g4Ú­4
0,
y=0
Ï §u =Ù­4Q 1 0 |f (x, y)| = x sin |x|
lim f (x, y) = 0
y
x→0
y→0
§ ¨ § 4ØQ§= ØQ lim lim f(x, y) = 0
= +∞
t→+0 t
1 + x2 + y2
lim
x→0
x2 + y2
= +∞
y→0
Ï (4) 0
sin(x3 + y3) x2 + y2
|x3 + y3| x2 + y2
u sin(x3 + y3)
lim
x→0
x2 + y2
=0
y→0
|x|3 + |y|3
|x|3
|y|3
=
+
x2 + y2 x2 + y2 x2 + y2
|xn − x0|2 + |yn − y0|2 |xn − x0| + |yn − y0|
q §u = xn → x0, yn → y0(n → ∞)
|xn − x0|2 + |yn − y0|2 → 0(n → ∞) (xn, yn) → (x0, y0)(n → ∞)
2. yy²²µµe#²lim¡þM'n X=M{0M§nb}Â#ñq§kuli§mMkn=M40 .
~µ QX 'g4 x − y
(1)
f (x, y) =
, x+y
x+y = 0
(0, 0)
0,
x+y = 0
x
−y
lim lim f (x, y) = lim = 1, lim lim f (x, y) = lim = −1
ux→0 y→0
x→0 x
y→0 x→0
lim lim f (x, y) = lim lim f (x, y).
x2 + y2
lim
=0
|x| + |y|
x→0
x→0 |x| + |y|
y→0
y→0
Ït
(2) lim √
§u √
= lim ( t + 1 + 1) = 2 lim
t→+0 t + 1 − 1 t→+0
x→0
y→0
x2 + y2 =2
x2 + y2 + 1 − 1
Ï §u 1 + t
(3) lim
'X ´ 'bFX + y2 = 1 (x, y) E
.
'S
X
¶
÷
vx2
+
y2
>
'X 1 (x, y)´E'©X¶¦Xθ9÷
(4) dknê9Ãnê'ÈS§&²¡þ¤kX(x,y)Ñ´E'bFX.
5. y²µeM0´²¡X8E'àX§uQE¥QXMn → M0(n → ∞).
y²µ®M0´²¡X8E'àX§δn Q 'X E M2, M2 = Mi(i = 0, 1)
n→∞
n→∞
d½Â§é §¨ §k ∀ε > 0, ∃N ∈ Z+
dqnØ1ª'§½k'üX§d '?¿S§& = M0,M0
r(M0, M0 ) ε
n>N
ε
ε
r(Mn, M0) < 2 , r(Mn, M0 ) < 2
r(Mn, M0) + r(Mn, M0 ) < ε
r(M0, M0 ) = 0 M0 = M0 .
(4) lim
x→0
x2 + y2
y→0
(5) lim (x2 + y2)e−(x+y)
x→+∞ y→+∞
ln(x + ey)
(6) lim
x→1 y→0
x2 + y2
)µ
Ï (1) 0
x2 + y2 |x| + |y|
§u (|x| + |y|)2 = |x| + |y| lim (|x| + |y|) = 0
x→a y→b
y→a
x→b
.
yk²µÏ­4Q§ué §¨ §ð |f(x, y) − A| < ε yQ ¥'½ § Qþª¥- §=& §ùÒy²
0 < |x − a| < δ
∀ε > 0, ∃δ > 0 |x − a| < δ, |y − b| < δ (x − a)2 + (y − b)2 = 0
ln(x + ey)
(6) lim
= ln 2
x→1 y→0
x2 + y2
207
3. Áye lim f(x, y) = AQ§ ¨x?Ûag§4lim f(x, y) = ϕ(x)Q§ug4Q§