April 20, 2006
Lecturer: Dr Martin Kurth
Trinity Term 2006
Problem Sheet 2 – Solutions
1. (a)
Z
∞
1
µ
¶
¶
Z ∞µ
1
1 −x
cosh x − ex dx =
e
dx
2
2
1
Z b
1
=
lim
e−x dx
2 b→∞ 1
¡
¢
1
= − lim e−b − e−1
2 b→∞
1
=
e
(b)
Z
1
∞
Z ∞
Z ∞
1
1
1+x
dx
=
dx
+
dx
2
x2
x
x
1
1
= ∞,
as the second integral diverges (see lecture).
(c) Substitute (eg)
u = x − 1,
which gives
Z
1
2
Z 1
1
1
dx =
du
x−1
u
0
= ∞
(see lecture)
(d) Following the hint, we see that
µ
¶
1
1
d
1
1
x sin
= sin − cos ,
dx
x
x x
x
and thus
¶
Z 2µ
1
1
1
sin − cos
dx =
x x
x
0
2
µ
1
1
1
lim
sin − cos
a→0+ a
x x
x
¯2
1¯
= lim x sin ¯¯
a→0+
x
Z
a
=
=
1
1
1
2 sin − lim a sin
2 a→0+
a
1
2 sin ,
2
¶
dx
as the second term is 0 due to the Sandwich Theorem with
−a ≤ a sin
1
≤ a.
a
2. (a)
lim
n→∞
4
=0
n2
(b)
n+3
lim
n→∞
n
µ
¶
3
= lim 1 +
n→∞
n
= 1+0
=
1
(c) Limit does not exist. Choose (eg) ² = 1/2. Then, for any real number
L and for any N , there exists an n > N with |an − L| ≥ 1 > ².
(d) Use the Sandwich Theorem with
−
1
1
1
≤ 2 sin(n) cos(n) ≤ 2
2
n
n
n
to get
lim
n→∞
µ
¶
1
sin(n) cos(n) = 0.
n2
3. Here we need the first two derivatives:
f 0 (x) = 4x3 − 2x
f 00 (x) = 12x2 − 2
Find the critical points c of f :
f 0 (c) = 0,
ie
2(2c2 − 1)c = 0,
√
√
giving us the critical points c = 0, c = 1/ 2 and c = −1/ 2. Plugging
these into the second derivative, we get
f 00 (0) = −2 < 0
and
√
√
f 00 (1/ 2) = f 00 (−1/ 2) = 4 > 0,
√
ie we have
√ a local maximum at c = 0 and local minima at c = 1/ 2 and
c = −1/ 2.
2
4. As f is continuous at L we know that for every ² > 0 there is a δ > 0 with
|x − L| < δ =⇒ |f (x) − f (L)| < ²,
and as limn→∞ an = L we know that there is an N such that
n > N =⇒ |an − L| < δ,
and thus
n > N =⇒ |f (an ) − f (L)| < ²,
which proves the statement.
3