华工-数学分析范例
时间：2025-06-26
《数学分析》中的一题多解汇编
（《数学分析》课程辅导材料—部分初稿）
《数学分析》课程组 编
华南理工大学 理学院 数学系
1
√n
n
n
,
Gn=
√n
(am+1Amm +11)
1/mm
=G.Am+1=
Am+AAm+1 Gm+1,
(1)
2
(1)
a1=a2=···=an.
An=
a1+a2+···+an
n 1
=An,
2
m
2
√n
n
n 1
a1+a2+···+an
,
a1+a2+···+an
n 1
a1+a2+···+an
n 1
a1a2
a3+a4+···+an
n
√n
n
=nlim→∞
an.
n 1
.
n 1
(3)
3
n
=lim
xn
yn yn 1
n→∞
=liman.
n→∞
2
δn=
a1+a2+···+an
n
n
A=|
k=1
n0
n0
n n0|(ak a)|+k=1
n0
1|(ak a)|+k=1
n
+ ,
(ak a)|.
n→∞
lim
A
n
n
=
sn0
n n0
(1
n0
n
+3M(1
n0
→0,1
n0
n
n
<
M
n
>
1
>M.sn
n→∞
lim
n
=+∞.
3.
f
[a,b]
c∈[a,b],
x∈[a,b],
f(c) f(x).
1
M={x∈[a,b]:f(x) f(u) u∈[a,x]}.
4
2(an
+bn).
u∈[an,cn]
v∈[cn,bn]
f(v) f(u),
an+1=an,bn+1=cn.
u∈[an,cn],
v∈[cn,bn],f(v)>f(u),
an+1=cn,bn+1=bn.
1
(i)an an+1 bn+1 bn.(ii)bn+1 an+1=
5
6
7
n}
1.
√2
n
1
n≥2√n
1 n·
√
0
√n
n
<<1+
2
,n
√
√n
2
√n
2
]
.
n 1< .
yn=
2
2
yn.
yn,
n>1
√n
]+1
.
2
2
6.
n→∞
lim
n
n!nn!
=e.
1
1.
ln(
Cauchy
nnlnn (ln2+ln3+···+lnn)=n!
n
.
n→∞
lim(bn+1 bn)=l lim
n)
bn
n→∞
=ln(1+
1
8
nn
√n
n!
,
n→∞
lim
an+1
(n+1)!
·
n!
n
)n=e.
7.
f
[a,b]
f
[a,b]
.
1(
)f
[a,b]
x∈[a,b],
O(x)
Mx,
|f(x)|<Mx
x∈O(x)
].
x∈[a,b]
(x)
OM
[a,bx,
[a,b]
.
O1,O2,···,On,
.
M1,M2,···,MnM=max{M1,M2,···,Mnn},
[a,b]
Oi,
i
=1x∈[a,b]
|f(x)|<M,f
[a,b]
.2(
Bolzano
.
f
[a,b]
)
a1=a,b1=b.
(a1+b1)
[a1,b1]
c1=
2n 1
(b a).
.
9
10
2
2)+
π
2
π
2
2
+
π
2
2
],m=[2Nπ+2π],([···]2N+
π
4
)
m>n>N,
4
,2Nπ+π<m<2Nπ+2π
√2
|sinn sinm|≥ε0=
.
|sinn|
3
n→∞
limsinn=A
sin(n+2) sinn=2sin1cos(n+1),
n→∞
lim2sin1cos(n+
n→∞
n→∞
1)=lim(sin(n+2) sinn)=A A=0,
n→∞√lim
n→∞
limcosn=0,A=limsinn=
11
√
k+1 2
√√
√k 2
(n+1)2
√
+1 2
√
√
(n+1)2
2
√
(n+1)2
√
≤lim
√
n+1,
1
n+1
n→∞
k=n2
2n+21≤k
1:(1)
x1=
√
(c+xn)
,(c>1
)
n→∞
limxn.c,
√√
c,f(x)=
c(1+x)
c+x
.c
xn>
c+xn
√
=f(xn)>f(
x1>
√
xn+1 xn=
c.
c(1+xn)
c+xn.
<0
{xn}
,x1<
.
√
c,xn
,xn
,
.
xn+1=
c(1+xn)
c.
2:
xn>0,
x>0
,f′(x)≡(c(1+x)
c2
(c+x)2
>0.
c>1
f′(x)=
c(c 1)
=1
1
c.
4.lim
f(x),g(x)
E
,x0
E
,
lim
n→x0
g(x)≤
n→x0
limf(x)+lim
12
f(xn)+lim(f(xn)+g(x))≤
g(xn).
.
5.
x
xlim
sinx tan→0
sinx
cosx 1
x3
=xlim
→0
cosx
2
.
2
sinx tanx
x3cosx
,
x(x→0),1 cosx～
12x2,
x( 1
1
xlim
sinx tanx
→0
x3cosx
=xlim→0x3
=
2)π,n
∈N+,
2
.
a
b
asinx+bcosx=
sinx～
a=0.
13
π
x′n=2nπ
n→∞,
2√
x
x=0
.
1(
Heine
)
xn=
1
,yn=
1
2
xn
=sin(2nπ+
π
yn
=sin2nπ=0,n∈N+.sin
1
yn
1
Heine
0.
x
x=0
2
.x′=
1
,x′′=
1
2
x3
.
1
3
L’Hospital
xlim
x tanx
→0
3x2=xlim
→0
2sec2xtanx=
1
3
1 sec2x
x3
=xlim
→0
3x2
=
1
x3sinx
.
sin
1
=
1
(1)
Heine
14
x3sinx
=lim
2sinx2
(3 x2)sinx+5xcosx
4cosx2 8x2sinx2
2 2
.
x→0
=lim
x→0
2
x→0
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