数学分析习题.pdf

1

2002


2002 12

≤
1
0 <b a.
(a − b)2 a + b √(a − b)2
≤− ab ≤.
8a 2 8b
√
−x 2 2

2
f(x)=ln(1+e ),g(x)= c + x (c>0)

3
sin x 2 π sin x √π
(1) 1 > ≥(0 <x≤); (2) > 3 cos x (0 <x≤);
x π 2 x 2
4 π 2 1 1
(3) (sin x)−2 ≤ x−2 +1−(0 <x≤); (4) < ln(1 + ) < √(x>0);
π2 2 2x +1 x x2 + x
ln x 1 4x πx πx
(5) ≤√(x>0,x= 1); (6) ≤ tan( ) ≤(0 ≤ x<1).
1 − x x π(1 − x2) 2 2(1 − x2)

5
(1 + x)α≥ 1+αx (x ≥−1, 0 <α<1); (1 + x)α≤ 1+αx (x ≥−1,α>1, or α<0).
+

∀∈
6
βα n N
1 1
(1 + )n+α≤ e ≤(1 + )n+β.
n n

n √√√

−
7
lim ( k +2 2 k +1+ k)
n→∞
k=1

8
(Stolz) ( ).
0 an − an+1

{ }
{ }

1 ( 0 ) an bn lim
n→∞ bn − bn+1
an an an − an+1

∞
∞
+ , lim + lim = lim .
n→∞ bn n→∞ bn n→∞ bn − bn+1
∞ an − an+1

{ }
∞

2 ( ∞) bn lim bn =+ . lim
n→∞ n→∞ bn − bn+1
an an an − an+1

∞
+∞,
lim + lim = lim .
n→∞ bn n→∞ bn n→∞ bn − bn+1
Hint of Proof
Only prove that conclusions (1) and (2) hold for the cases with the finite limit, the
proof of the other cases are similar.
an − an+1 +
The proof of (1). Suppose that lim = l ∈ R. Then, ∀ε>0, there exists N1 ∈ N such
n→∞ bn − bn+1
that
an − an+1
| − l| <ε
bn − bn+1
hold for all n ≥ N1. Furth