高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类
ε>0
2x+32
<ε3x3
x
X
δ
X
δ
ε
f(x) A=
2x+321
=<ε,3x3x
x>
1
ε
X=
1
ε
x>X
2x+32
<ε3x32x+32
=
x→+∞3x3lim
ε
f(x) A=
1
sinxx
0≤
1x
1x
<ε
1
x>
sinxx
ε2
0<ε
X=
ε2
x>X
x>0
sinxx
0<ε
x→+∞
lim
sinxx
=0
f(x) A=1
2
1x 2
, 1=
x 1x 1
0<x 2<δ
δ
x
x 2<
x→x0
f(x)
x0
x
f(x)
1
x 2<
2
35<x<22
ε
2
x 2<ε31 2
x 22x 2<=x 2x 133
x 2<ε
2
δ=min ε,
f(x) A=
3 2
0<x 2<δ
x 22
<x 2x 13
2
x 2<ε3
x 22
<x 2<ε,x 13
lim
x→2
1
=1x 1x <
12
x2 1x 1
f(x) A=2 2=
xx x
13<x<22
x 1x 1
<=2x 1x