Solutions to homework 9
11.3.3(a)(b)
a) For any x > 0 and 0 ≤ t ≤ π
sin t
sin t
≤
≤ sin t.
1 + πx
1 + xt
Integrating the above inequalities
2
≤
1 + πx
Since
x→0
lim+
x→0
b) We need to show
lim
x→0+
0
1
0
π
0
sin tdt
= 2.
1 + xt
t2
dt = lim
1 + t4 x
x→0−
The latter integral equals
1
0
sin tdt
≤ 2.
1 + xt
2
= 2,
1 + πx
lim+
by the Squeeze theorem
π
0
1
t2
dt =
1 + t4 x
1
0
t2 dt = 1/3.
For the first limit we have that for any x > 0, and 0 ≤ t ≤ 1
t2
t2
≤
≤ t2 .
1+x
1 + t4 x
Integrating
1
≤
3(1 + x)
Since
lim
x→0+
0
1
1
t2
dt ≤ .
1 + t4 x
3
1
1
= ,
3(1 + x)
3
the Squeeze theorem gives
1
lim
x→0+
0
1
t2
dt = .
1 + t4 x
3
1
t2 dt
For the second limit we have that for any x < 0, and 0 ≤ t ≤ 1
t2 ≤
Integrating
1
≤
3
1
0
t2
t2
≤
.
4
1+t x
1+x
1
t2
dt ≤
.
4
1+t x
3(1 + x)
Since
lim
x→0−
1
1
= ,
3(1 + x)
3
the Squeeze theorem gives
lim−
x→0
0
1
1
t2
dt = .
1 + t4 x
3
This implies continuity at 0.
11.5.4
Using theorem 2.5 we know that for any a ∈ R and any δ-neighborhood of a:
(a − δ, a + δ), there exists a rational number q ∈ Q and an irrational number
s ∈ R, s ∈ R such that both these numbers belong to the δ-neighborhood.
Therefore for any ε > 0, ε < 1 and any δ we have for a ∈ Q
|f (a) − f (s)| = 1 > ε, |a − s| < δ.
or a ∈ Q
|f (a) − f (q)| = 1 > ε, |a − s| < δ.
Hence f (x) is not continuous at any a.
2